MATRICULA TION

STANDARD X

Untouchability is a sin

Untouchability is a crime

Untouchability is inhuman

PHYSICPHYSICPHYSICPHYSICPHYSICSSSSS

c Government of TamilnaduFirst Edition - 2006

CHAIRPERSON

Dr. N. VIJAYANCorrespondent & Principal

Zion Matriculation Hr. Sec. School Indira Nagar, Selaiyur, Chennai - 600 073.

REVIEWERS

Dr. V. MIRASU Thiru. G. VIJAYAKUMARPrincipal, Principal,St. John’s Mat. Hr. Sec. School Jaigopal Garodia Mat.Hr. Sec. SchoolBaba Nagar, Villivakkam, S.R.P. Colony,Chennai - 600 049. Chennai - 600 082.

AUTHORS

Tmt. E. DORIS JAMES Tmt. AGNES SAMUELP.G. Asst. in Physics, (Retd.) B.T. Asst,Sacred Heart Mat. Hr. Sec. School C.S.I. Ewart Mat. Hr. Sec. SchoolChurch Park, Chennai - 600 084.Chennai - 600 006.

Price: Rs.

The book has been prepared by the Directorate of Matriculationschools on behalf of the Government of Tamilnadu.

This book has been printed on 70 GSM paper.

Printed by Offset at :

PREFACE

As a teacher of Physics, I have always found myself inspired bythe words of Professor D.S. Kothari, a renowned Physicist whooften said to his students, “You must regard yourself fortunate thatyou have got a chance to study Physics, for Physics a fundamentaland exciting science, is the basis of all sciences”. This book has beenmeticulously designed to provide a seamless transition to the HigherSecondary course. The enriched contents will accelerate the desire toprobe into the concepts. The challenging presentation of physical ideas aimat developing the spirit of enquiry, to enhance creativity and aesthetic sense.Each concept has been explained clearly, in simple and lucid language in order toamplify the students’ scientific temper for problem solving and critical thinking.The systematic and progressive explanations and illustrations, evolve a logicalapproach and rational analysis, enabling the students to master the essenceof science in general and physics in particular.

The salient features of this book are

* The thorough explanation of concepts lead the students gradually and methodically from the known to the unknown.

* The concepts explained are based on daily life situation.

* ‘Let us muse upon’ - provided at the end of each chapter serve as a ready reckoner for quick revision.

* Every chapter has a number of thought provoking questions.

The information given under ‘Know it yourself ’ is only to enhance the knowledge of the students and not for evaluation.

Dr. N. VijayanChairperson

iii

ACKNOWLEGEMENT

I gratefully acknowledge the cheerful and constant

help rendered by Mr. S.Thapasi, PGT in Physics,

Wesley Higher Secondary School, Royapettah,

Chennai Mr. R. A. Balaraman, PGT. Physics and

Mr. E. Sampath Kumar, TGT. Physics, St. John’s

Matriculation Hr. Sec. School, Villivakkam, Chennai.

I am most indebted to Dr. K. Prabhakar Rao,

Professor in Physics, Osmania University, Hyderabad

and Prof. M.V. Bhagwat, H.O.D Physics, St. Xavier’s

College, Mumbai for their expert opinion.

I sincerly appreciate Mr. T. Magesh Rajan,

Computer Software Designer of my school for

his innovative and skillful workmanship.

iv

CHAPTER PAGE

1. MECHANICS 1

2. LIGHT 21

3. ELECTRICITY 48

4. PROPERTIES OF MATTER 74

5. MODERN PHYSICS 94

6. X - RAYS AND RADIOACTIVITY 113

7. UNIVERSE 126

PRACTICAL 138

MODEL QUESTION PAPER 161

CONTENTS

v

(PRACTICAL)

1

1. MECHANICS

1.1 Centre of gravity

All objects are made up of tiny particles. Each particle has a mass ofnegligible volume. These particles behave as if their mass is concentrated atsome point. The point at which their mass appears to act or is concentratedis called its ‘centre of mass’. When the shape of the object is regular, thenthe centre of mass is its geometric centre.

When an object falls towards the earth, every particle of the object is pulled bythe force of gravity and the object behaves as if all that force is concentrated atone point in the object. This point is called the ‘centre of gravity’

The centre of gravity of an object is a fixed point through which theentire weight of the object acts, irrespective of the position of the object.

‘Joy in looking and comprehending is nature’s most beautiful gift’, said AlbertEinstein.

In the world around us, we see a falling apple, an orbiting satellite, a speedingrace car, an acrobat balancing on a bar. This is what ‘Mechanics’ is all about.

Mechanics is that branch of physics which deals with the conditions of rest ormotion of bodies around us. Statics is that branch of mechanics which deals withthe science of forces balancing one another. Dynamics is that branch of mechanicswhich deals with the motion of bodies under the action of forces.

The falling apple reminds us of Sir Isaac Newton. He was Einstein’s predecessorin understanding gravity. His theories of gravity help us to explain the fall of anapple or the path of a satellite. Albert Einstien represents the human spirit andcreative drive in all of us. Now it is your turn to experience the excitement andchallenge of Physics.

In this chapter, we shall study about the fundamental concepts of centre ofgravity, projectile motion and circular motion.

2

3. Circular ring, disc, At its geometricsolid sphere, hollow centre.sphere.

4. Triangle.

At the point ofintersection of themedians. This pointis called the centroid.

5. Right cylinder.

At the midpoint of itsaxis, h/2 from the basewhere ‘h’ is the heightof the cylinder.

6. Solid cone.At a point on the axis, h/4 from the basewhere ‘h’ is the heightof the cone.

1.2 Illustrations for the stability of bodies

(a) A man carrying a bucket of water in his right hand, leans towards his left.He does it to keep the vertical line through the centre of gravity fall between hislegs.

(b) The bottom of a ship is made heavy, to keep the centre of gravity as low aspossible. The cargo is kept at the base of the ship and this makes the ship stable.

CG

CG

CG

1. Uniform rod. At the midpoint ofthe rod.

2. Rectangle.At the point ofintersection of thediagonals.

TABLE 1.1 Centre of gravity of regular objects

No. Name of the objectFigure denoting the

position of thecentre of gravity

Position of thecentre of gravity

of the object

3

(c) In the ‘Leaning Tower’ of Pisa in Italy, thevertical line through the centre of gravity falls withinthe base of the tower. Thus the tower does notfall and remains stable. (Fig. 1.1).

(d) A student carrying a school bag andclimbing up a flight of stairs always leans forwardto maintain his stability.

(e) ‘Tanjore - dolls’ and ‘Hit-me’ dolls have abroad and heavy base to enable them to comeback to a vertical position when they are knockeddown. (Fig. 1.2).

Conditions for the stability of bodiesFrom the above illustrations it is understood

that the following conditions are necessary for anobject to be stable.

(i) The base of the body should be broad.(ii) The centre of gravity should be as low as possible.(iii) The vertical line through the centre of gravity should fall within the base.

1.3 Free fallGalileo stated “In the absence of air resistance, all bodies fall at the same

acceleration, which is approximately 9.8m s–2 ”.

The value of acceleration due to gravity at different locations are given below.

TABLE 1.2

Fig. 1.2 Tanjore doll.

Fig. 1.1 The leaning tower of Pisa

Location Acceleration due to gravity (ms-2)

1. At the equator 9.7802. At the poles 9.8323. At Chennai 9.7824. At the surface of the earth (average) 9.85. At the centre of the earth zero

4

An object is said to have a free fall when it falls vertically under theinfluence of gravity alone, free from air resistance.

Examples

(i) Fruits like ripe mangoes falling from a tree.(ii) A skydiver with an unopened parachute.(iii) Bungee jumper.

Imagine a body falling from a certainheight. In each succeeding second of itsfall, it is seen that the velocity of thebody increases by 9.8 ms–1. Initiallythe velocity of the body is zero . As itdescends down, the velocity increases andreaches a maximum value on reaching theground.

1.3.1 Newton’s feather and coin experiment

A dry leaf, a feather or a sheet of papermay flutter to the ground, while, a stoneor a coin falls rapidly. The fact that airresistance is responsible for these differentaccelerations is explained with the‘Feather and coin’ experiment performedby Sir Isaac Newton.

Newton took a long glass tube asshown in Fig. 1.3 and dropped a featherand a coin simultaneously into it. Henoticed that the coin travelled much fasterthan the feather because of the air insidethe tube.

Galileo Galilei

Fig. 1.3 Feather and coinexperiment

5

Then the air from the glass tube was removed with the help of a vacuum pump.When the experiment was repeated, the coin and the feather reached the otherend of the tube at the same time.

From this experiment he proved that in the absence of air, all objects fall withthe same acceleration. Thus Galileo’s statement was proved by Newton.

The velocity of a freely falling object (under

the influence of the force of gravity) increases at a

constant rate. Hence, it is said to be accelerated. This

acceleration is known as ‘acceleration due to gravity’

and it is denoted as ‘g’. The average value of ‘g’ has

been found to be 9.8 ms–2.

1.3.2 Equations of motion for a freely falling object

When an object is dropped from a certain height

initial velocity u = 0

acceleration due to gravity a= + g

displacement s = h

v = u + at becomes v = gt

s = ut + at2 becomes h = gt2

v2 = u2 + 2 as becomesv2 = 2gh

When an object is thrown vertically up,

initial velocity = u u = gt

acceleration due to gravity a= –g h = ut – gt2

displacements = h u2 = 2 gh

final velocity v = 0 at the maximum height

12

12

12

SirIsaac Newton

6

Problems

1. A cat steps off a ledge and falls to the ground in 2 seconds

(i) what is its velocity when it reaches the ground?

(ii) What is its average velocity?

time taken t = 2 s

acceleration due to gravity g = 9.8m s–2

(i) velocity v = gt = 9.8 × 2 = 19.6 ms–1

(ii) average velocity = initial velocity + final velocity2

= 0 + 19.6 = 9.8 ms–1

2

2. An object released from a certain height, hits the ground with a velocity of49m s–1. What is the height from which it is released?

initial velocity u = 0

final velocity v = 49m s–1

acceleration due to gravity g= 9.8m s–2

v2 = 2gh

49 x 49 = 2 × 9.8 × h

∴ h = 49 × 492 × 9.8

h = 122.5 m

7

3. A ball is thrown up with a velocity of 19.6m s–1. Find the maximum heightreached and the time taken to reach it.

initial velocity u = 19.6 ms–1

final velocity v = 0

acceleration due to gravityg = 9.8 ms–2

a = –g

u2 = 2 gh

we know that h = u2 = 19.6 × 19.6 = 19.6 m2g 2 × 9.8

h = 19.6 m

we know that u = gt

t = u = 19.6 = 2 sg 9.8

1.4 Projectile motion

An object thrown with an initial velocity horizontally or at an angle lessthan 900 under the action of the earth’ s gravity is called projectile.

A few examples of projectile are (i)a bullet fired from a gun

(ii) a stone thrown horizontally from the top of a building (iii) a bomb dropped

from an aeroplane and (iv) a shotput or a javelin thrown by a sportsman.

What is the path taken by a coin when it is pushed

horizontally off the edge of a table? (as shown in Fig

1.4). It takes a curved path undergoing both horizontal

and vertical motion.

Consider a body A which is allowed to fall freely

and another body B which is projected horizontally

with a velocity u from the same height and at the same Fig. 1.4 coin fallingfrom a table

8

time as shown in Fig. 1.5. The body A which is falling

freely and the body B which is projected horizontally

from the same height and at the same time will reach

the ground simultaneously, at different points. The two

bodies at any instant will be at the same vertical height

above the ground.

Range of a projectile is the horizontal distance between the point ofprojection and the point where the projectile hits the ground.

1.4.1 Path of a projectile

Consider an object thrown with a horizontal velocity‘u’ from a point P which is at a certain height fromthe ground as shown in Fig. 1.6. As it undergoesvertical and horizontal motion, it has a verticaland horizontal velocity.

The vertical velocity increases due to the force ofgravity. But the horizontal velocity remains constant throughout its motion.

Let us assume that the object moves from P to O in a time interval ‘t’.

Horizontal displacement of the particle x = ut ———(1)

the vertical displacement of the particle y = gt2 ——(2)

from equation (1) t =

substituting this in equation (2) y = gt2

y = g

y = . x2

y = k . x2 where k =

(since u and g are constants k is also a constant)

This is the equation of a parabola. Thus, the path of a projectile is a parabola.

12

xu

12

xu( )

2 x2

u2

g2u2

g2u2

12

12

Fig. 1.5 Free fall and projectile motion

Fig. 1.6 Path of a projectile

= g .

Since the initial vertical velocity is zero,

9

Problem

1. A body is thrown horizontally from the top of a building. It falls on theground after 1.5 s at a distance of 5m from the building. Calculate(a) the height of the building (b) the horizontal velocity?

time taken t = 1.5srange x = 5 m

acceleration due to gravity g= 9.8m s–2

h = gt2

h = × 9.8 × 1.5 × 1.5

height of the building h = 11.025mrange = horizontal velocity × time takeni.e. x = u t

horizontal velocity u = = = 3.33m s–1

2. An aeroplane, flying horizontally at a height of 1960m with avelocity of 125m s–1 aims to hit an enemy target. Find at what distance from thetank, the pilot will release the bomb in order to hit the tank.

acceleration due to gravity, g = 9.8m s–2

height from the ground, h= 1960m

Since the initial vertical velocity is zero,

h = gt2

1960 = × 9.8 × t2

t2 = = 400

∴ t = 20 s.

horizontal distance = horizontal velocity × time taken

= 125 × 20 = 2500m.

The bomb should be released at a distance of 2500 m from the tank.

12

xt

51.5

1212

1219604.9

10

1.5 Types of motion

There are different types of motion. They are(i) linear or translatory motion. (ii) vibratory motion(iii) circular motion (iv) random motion and(v) oscillatory motion

A child sliding down a sliding board, an aeroplane on the runway, aspeeding train, a man walking down a straight road are examples of objectswhich move in a straight line. This is called linear motion.

Circular motion

(i) Rotatory motion The rotating blades of a fan, a potter’s wheel, thespinning earth, a merry-go-round and a giant wheel move about their axes. Such amotion is called rotatory motion.

(ii) Revolving motion The motion of the moon round the earth, themotion of an electron around the nucleus are examples of ‘revolving motion’.

The movement of a particle in a circular path is called circularmotion.

1.5.1 Angular displacement and angular velocity

Let us consider an object moving ina circular path with uniform speed abouta fixed point O as centre. When theobject moves from A to B, the radius OAsweeps an angle θ at the centre as shownin Fig. 1.7.

Angular displacement of the object is measured by the angleswept by the radius at the centre, as the particle is moving along thecircumference of the circle.

The unit of angular displacement is radian. One radian is the anglesubtended at the centre of a circle by an arc whose length is equal to theradius of the circle.

Fig. 1.7 Circular motion

11

Angular velocity of an object is the rate of change of angulardisplacement.

The object moves from A to B in time ‘t’. During this time the radius sweeps anangle θ at the centre as shown in Fig. 1.7.

angular velocity, ω =

ω =

The unit of angular velocity is radians per second. (rad s–1)

Angular acceleration is the rate of change of angular velocity.

Angular acceleration, α =

α =

where ωo and ω are the initial and final angular velocities in a time

interval ‘t’. The unit of angular acceleration is rad s–2.

1.5.2. Relation between linear velocity and angular velocity

Consider a body moving along the circumference of a circle of radius ‘r’ withlinear velocity ‘v’ and angular velocity ‘ω’. It covers AB in a time interval ‘t’.

The linear velocity of the body v = = -------(1)

s = r θ -----(2) (one radian corresponds to r.∴ θ radian corresponds to rθ)

substituting equation (2) in (1)

v = = -------(3)

v = r ω, since ω =

linear velocity = radius of the circle ××××× angular velocity

angular displacementtime

θ

t

Change in angular velocitytime

ω — ωo

t

θ

t

st

rθt

Fig. 1.8 Circular motion

ABt

st

12

1.5.3. Angular momentum

Angular momentum of a particle is defined as the moment of linearmomentum of the particle.

The moment of linear momentum= mv × r

= mrω × r, ... v = rω= mr2ω= I ω

where I = mr2 is called the moment of inertia of the rotating particle.Moment of inertia is the measure of inertia of the particle in circularmotion and it is equal to the product of the mass of the particle and thesquare of its distance from the axis of rotation.

∴ Angular momentum, L = Iω

Angular momentum = moment of inertia ××××× angular velocity

The unit of angular momentum is kgm2s–1

TABLE 1.3. Comparison of linear and angular motion

No. Linear motion Angular motion

1. Linear displacement is ‘s’ Angular displacement is ‘θ’

2. Linear velocity, v = Angular velocity, ω =

3. Linear acceleration, a = Angular acceleration, α =

4. Mass ‘m’ is the measure of Moment of inertia I = mr2, isinertia in linear motion the measure of inertia in

circular motion.

5. Linear momentum, p = mv Angular momentum, L = Iω

st

θ

tv – u

tω – ω

o

t

( )

13

1.6 Centripetal force

A stone is tied to a string and rotated in acircular path. The pull of our hand on the stringis towards the centre of the circular path tracedby the stone. The tension in the string providesthe centripetal force. If the string is suddenlyreleased, the stone will go off in a directiontangential to the circle.

When a satellite moves in a circular path around the earth, the gravitational

force between the earth and the satellite provides the necessary

centripetal force. For a car taking a curve, the frictional force on the tyres and the

ground provides the required centripetal force.

The external force required to make a body move along a circularpath with uniform speed and directed towards the centre is calledcentripetal force.

Let us consider an object of mass ‘m’, moving along a circular path of radius‘r’

with an angular velocity ω and linear velocity ‘v’. As the body moves along the

circular path, the direction of velocity changes at every point and hence it has an

acceleration, which is directed towards the centre. This acceleration is called

centripetal acceleration.

Centripetal force F = mass × acceleration

=

Centripetal forceF = m r ωωωωω2 ... v = rω

m × v2

r

Fig. 1.9

a = v2

r

( )

Centripetal acceleration

= mv2

r

14

Problem

1. A particle undergoing circular motion of radius 7 cm completes onerevolution in 0.2s. Calculate its angular velocity and linear velocity.

The angle swept in onerevolution θ = 2π radian

radius of the circle r = 7cm = 0.07 m

time taken t = 0.2s

angular velocity ω = = = 10 π rad s–1

linear velocity v = rω

v = 0.07 × 10π = 0.07 x 10 x 22

linear velocity v = 2.2 ms–1

2. A body of mass 2kg undergoes circular motion in a path of radius 1.4mwith an angular velocity of 3.14 rad s-1. Calculate its centripetal acceleration andcentripetal force.

mass of the body m= 2kg

radius of the circle r = 1.4m

angular velocity ω = 3.14rad s–1

centripetal acceleration a= rω2

= 1.4 × 3.14 × 3.14

= 13.8 ms–2

centripetal force F = mrω2

= 2 × 1.4 × 3.14 × 3.14

centripetal force F = 27.60 N

θ

t2π

0.2

7

15

3. A giant wheel rotates with a linear velocity of 20 ms–1 at a point on itscircumference. If the diameter of the wheel is 100 m, what is the centripetalacceleration at that point?

diameter of the wheel= 100m

radius r = 50m

velocity = 20m s–1

centripetal acceleration a= =

centripetal acceleration= 8 m s–2

4. Calculate the centripetal force on a 1 kg rock whirling at a speed of5 m s–1 in an arc of radius 1.5 m.

mass of the rock = 1 kg

speed of the rock = 5 m s–1

radius of the arc = 1.5 m

centripetal force F =

=

=

centripetal forceF = 16.67 N.

v2

r20 × 20

50

mv2

r

1 × (5)2

1.5

1 × 5 × 5

1.5

16

Let us muse upon

F The study of moving objects under the action of force is called dynamicsand the study of forces on bodies at rest is called statics.

F The centre of gravity of an object is a fixed point through which theentire weight of the object acts, irrespective of the position of the object.

F The acceleration due to gravity decreases with increasing altitude andit decreases with depth.

F The stability of a body is decided by the position of the centre ofgravity. A body will be stable if the vertical line through the centre ofgravity falls within the base

F A projectile is defined as a body which is given an initial velocityand then allowed to move under the action of gravity.

F The Range of a projectile is the horizontal distance between thepoint of projection and the point where the projectile hits the ground.

F The movement of a particle in a circular path is called circular motion.

F Angular displacement of an object is measured by the angle swept by theradius at the centre as the particle moves along the circumference of a circle

F The unit of angular displacement is radian. One radian is defined as theangle subtended at the centre of the circle by an arc whose length isequal to its radius.

F Angular velocity of an object is the rate of change of angular displacement.

F The relation between linear and angular velocity is v = rω.F Angular momentum of a particle is defined as the moment of linear

momentum of the particle about the axis of rotation.

F The moment of inertia is defined as the measure of inertia of arotating particle and is equal to the product of the mass of theparticle and the square of its distance from the axis of rotation.

F The external force required to make a body move along a circular path withuniform speed and directed towards its centre is called centripetal force.

F When a body moves along a circular path, the direction of velocitychanges at every point and hence it has an acceleration which is directedtowards the centre. This acceleration is called centripetal acceleration.

17

Self evaluation

1.1. The centre of gravity of a triangular body lies at its

(a) centroid (b) orthocentre (c) vertex (d) circumcentre

1.2. The average value of acceleration due to gravity is

(a) 9.8ms–2 (b) 9.8cms-2 (c) 19.8ms–2 (d) 8.9ms–2

1.3. A man carrying a cement bag on his back up a slope will

(a) lean backward (b) lean forward (c) walk straight

(d) lean towards his left.

1.4. When an object falls vertically the initial velocity is

(a) more than 9.8 ms–1 (b) less than 4.6m s–2 (c) 2m s–1

(d) zero.

1.5. A ball is thrown vertically up, then the velocity at its maximum height is

(a) more than 9.8 ms–1 (b) zero (c) less than 9.8 ms–2

(d) 9.8 ms–1

1.6. The force between the moon and the earth which acts as a centripetal forcekeeping the moon in its orbit is

(a) an elastic force (b) gravitational force (c) electrical force

(d) nuclear force.

1.7. An object released from a height of 19.6 m strikes the ground with avelocity of

(a) 7 ms–1 (b) 9.8m s–1 (c) 19.6m s–1 (d) 15m s–1

1.8. A marble is thrown up vertically with an initial velocity of 9.8 ms–1. Themaximum height reached is

(a) 9.8 m (b) 8.9 m (c) 4.9 m (d) 19.6 m

1.9. The angular velocity of the seconds hand of a wrist-watch is

(a) rad s–1 (b) rad s–1 (c) rad s–1 (d) rad s–1

1.10. A cycle wheel 0.6m radius is moving with an angular velocity of10 rad s–1. Its linear velocity is

(a) 2m s–1 (b) 60m s–1 (c) 10 ms–1 (d) 6 ms–1

π

15π

30π

60π

90

18

1.11. What is centre of mass?

1.12. Define the term centre of gravity.

1.13. Where does the centre of gravity of a rectangular cardboard lie?

1.14. At what position does the centre of gravity of a cylindrical tin lie?

1.15. Mention the conditions for the stability of bodies.

1.16. When do we say an object is in free fall? Mention some examples.

1.17. With the help of a diagram explain Newton’s ‘Feather and coin’ experiment.

1.18. Write the equations of motion for(a) a freely falling object (b) an object thrown vertically up.

1.19. Define a projectile. Give some examples of projectile.

1.20. Why does a ball released from a moving bus take a parabolic path?

1.21. Define the term range of a projectile.

1.22. A boy throws a water-balloon from a certain height on a passerby.What is the path taken by the water-balloon?

1.23. Show that the path of a projectile is a parabola.

1.24. What is circular motion? Give some examples.

1.25. Define angular displacement and mention its unit.

1.26. What is angular velocity? Give its formula and unit.

1.27. Define angular acceleration. Write its formula and unit.

1.28. Derive the relation between linear velocity and angular velocity.

1.29. Define angular momentum and derive its formula.

1.30. Tabulate the differences between linear and angular motion.

1.31. Give two examples of objects moving under the influence of centripetalforce and hence define it.

1.32. What is centripetal acceleration? Give an expression for centripetal force.

1.33. What are the different types of motion?

1.34. Define moment of inertia.

19

1.35. Compare the motion of a freely falling body with that of a projectile.

1.36. An aeroplane flying horizontally with a velocity of 100 ms–1, releasesa bomb at a height of 490 m. Find the time taken by the bomb to reach theground.

1.37. A marble is thrown up vertically with an initial velocity of 29.4 m s–1.How high will it rise and how long will it take to reach that height ?

1.38. Calculate the angular velocity of the earth about its own axis.

1.39. A fighter pilot flies the aircraft in a circular path of radius 200 m. If theaircraft flies with the speed of 400 km hr–1 determine the centripetalacceleration.

1.40. A force of 150 N is required to break a 3m long nylon cord. An object ofmass 1.2 kg is fixed to one end of the cord and whirled around. Determinethe maximum speed with which it can be whirled around withoutbreaking the cord.

1.41. What speed should the rim of a 50 m diameter space station travel, so that itsinhabitants experience an acceleration of 10 m s–2 ?

1.42. A body is dropped from a certain height. What is the velocity of the body atthe third second of its fall ?

1.43. A ball is released from a certain height. Calculate the vertical distancecovered by the ball at the sixth second of its fall.

1.44. To estimate the height of a bridge across the river, a stone is dropped freelyin the river from the bridge. The stone takes 2s to touch the water surface.Calculate the height of the bridge from the water level.

1.45. A merry-go-round completes 3 revolutions in one minute. Calculate itsangular velocity.

1.46. The centripetal force acting on a body of mass 5 kg, undergoingcircular motion of radius 1m is 500 N. Calculate (a) its linear velocity (b) itsangular velocity.

1.47. A soft ball is dropped from a height of 10m. How long does it take for the ballto reach the ground?

20

1.48. A toy car is dropped from the top of a building. It reaches the groundin 3 s. Calculate (a) the velocity with which it strikes the ground(b) the height of the building.

Answers

1.1. (a) 1.2. (a) 1.3. (b) 1.4. (d) 1.5. (b)

1.6. (b) 1.7. (c) 1.8.(c) 1.9.(b) 1.10.(d)

1.36. 10 s. 1.37. 44.1m, 3s.

1.38. rad s–1. 1.39. 61.73 m s–2

1.40. 19.36 m s–1. 1.41.15.8 m s–1.

1.42. 29.4 m s–1. 1.43. 176.4 m.

1.44. 19.6 m. 1.45. 0.314 rad s–1.

1.46. 10 m s–1, 10 rad s–1. 1.47. 1.43 s.

1.48. 29.4 m s–1, 44.1m.

π

43200

21

2. LIGHT

In one of the greatest theoretical developments of the nineteenth century,James Clark Maxwell predicted that the change in both electric and magneticfields cause electro magnetic disturbancesthat travel through space. It is similar tothe spreading water waves created by apebble thrown into a pool. Thedisturbance caused by electric andmagnetic fields propagates in adirection perpendicular to both theelectric and the magnetic fields asshown in Fig. 2.1

Oscillating electric and magnetic fields perpendicular to each othercause disturbances in space. These disturbances propagate in adirection perpendicular to both the electric and the magnetic fieldsand are called electromagnetic waves.

They are transverse in nature and do not require a material medium forpropagation. They transfer energy from one point to another.

Two decades after Maxwell’s prediction in 1888, Heinrich Hertzproduced electromagnetic waves by means of oscillating electric charges.Moreover he conducted numerous experiments to demonstrate that theybehaved like light waves, satisfying all the phenomena of light.

Fig . 2.1 Electromagnetic waves

Y

Z

E B E B

X

B E B E

James Clark Maxwell

22

The inference from Hertz experiments

was that light waves are electromagnetic

waves. They are the visible part of the

electromagnetic spectrum and occupy a

very small band in it.

Velocity of electromagnetic waves

Maxwell arrived at an equation to findthe value of the velocity of electro-magnetic waves. It is given by

where µο is the permeability of free space and εο is the permittivity offree space. The value of velocity of electromagnetic waves is found to bec = 3 × 108 m s–1

2.1 Electromagnetic spectrum

Electromagnetic waves in the range of wavelength from a few kilometre downto 10–14m constitute the electromagnetic spectrum. The corresponding range offrequency is in the order of a few hertz to 1022 Hz.

Classification of electromagnetic spectrum

The electromagnetic spectrum is classified as

(i) Gamma rays

(ii) X-rays

(iii) Ultra-violet rays

(iv) Visible light

(v) Infra-red rays

(vi) Microwaves and

(vii) Radio waves

Hertz1

c =√ µο εο

23

Brief descriptions of the electromagnetic spectrum are given below in theincreasing order of wavelength.

Gamma rays These are emitted by radioactive nuclei and have wavelengthbetween 10–14m and 10–10m. They are highly penetrating and cause seriousdamage if absorbed by living tissues. Those working near such radiationsmust be protected using aprons made of lead.

X-rays They were discovered by Roentgen in 1895. X-rays are producedwhen fast moving electrons are suddenly stopped by elements of highmass number like molybdenum or tungsten. Their wavelength ranges from10–10m to 3× 10–8 m. X – rays are used for diagnostic purposes

Ultra-violet rays Their wavelength ranges from 6×10–10m to 3.8×10–7m. Sunis a natural source of ultraviolet rays. Most of these rays are absorbed by theozone layer in the stratosphere.

Visible light This is the most familiar form of electromagnetic waves.It is emitted by incandescent solids, sun, stars and fluorescent lamps. Thewavelength range is from 3.8×10–7m to 7.8×10–7m and it consists of seven coloursnamely Violet, Indigo, Blue, Green, Yellow, Orange and Red (VIBGYOR)

Table 2.1 Wavelength range of different colours

Colour Wavelength A (10–10m)

Violet 3800 — 4400

Indigo 4400 — 4600

Blue 4600 — 5000

Green 5000 — 5700

Yellow 5700 — 5900

Orange 5900 — 6200

Red 6200 — 7800

o

24

Infra-red rays Their wavelength range is from 7.8× 10–7m to 3× 10–5m.They are used in physiotherapy for the treatment of arthritis.

Micro waves Their wavelength ranges from 10–3m to 0.3m. They aregenerated by electronic devices like magnetron, klystron and travelling wave tube.They are used in telegraph, telephone, television and radar communicationsystems and also in micro wave ovens.

Radio waves They are produced by accelerating electric charges throughconducting wires. They are used in radio and television transmission and cell phones.Their wavelength range is from 10m to 104m.

The frequency range of radio waves goes up to 109 Hz.FM band (frequency modulation) is from 88MHz to 108MHz.AM band (amplitude modulation) ranges from 550kHz to 1600kHz.Television occupies frequency bands on either side of the FM region. Theallocation of frequency ranges for various broadcasting stations is done at theinternational level.

Know it yourself

1888 — Hertz produced electro magnetic wavesof wavelength about 6m.

1895 — J.C. Bose, an Indian scientist producedelectromagnetic waves of wave lengthranging from 5 mm to 25 mm.

1899 — G. Marconi was the first to transmitelectromagnetic waves and establishwireless communication across theEnglish Channel.

25

Problem

1. Calculate the frequency of a wave of wave length 150m.

wavelength λ = 150m

velocity c = 3 × 108 m s–1

c = υλ

3 × 108 = υ × 150

frequency υ = 3 × 108 Hz 150

= 2 × 106 Hz

frequency = 2 MHz.

2.What is the wavelength range covered by the AM radio band withfrequencies in the range of 550 kHz to 1600 kHz?

Frequency range = 550kHz to 1600kHz

= 550 × 103 Hz to 1600 × 103 Hz.

c = 3 × 108 m s–1

Wavelength λ = cυ

∴ Wavelength range = 3 × 108 m to 3 × 108 m

550 × 103 1600 × 103

Wavelength range = 545.5 m to 187.5 m

2.2 The visible spectrum

For centuries, it had been known that glass and precious stones glitteredin bright colours when white light is passed through them. In the middle of theseventeenth century, Newton investigated this systematically.

26

Newton’s experiment

Newton made a small circular hole in one of the window shutters of his room inCambridge. When the sun’s rays passed through this hole, they made a circularpatch on the opposite wall. On placing a prism in front of the hole an elongatedcoloured patch of light was seen on the wall.

Newton called this coloured patch as spectrum and noted that the colours werein the order of Violet, Indigo, Blue, Greeen, Yellow, Orange and Red as shown inFig. 2.2.

The splitting of white or any composite light into its constituent colourson passing through a prism is called dispersion.

Refraction through a prism

When a ray of light passes through a prism, it is deviated from its initialpath as shown in Fig. 2.3. The angle between the incident ray and the

Fig. 2.2. Newton’s experiment with a prism

Sun light

Window

shutter

VIBGYOR

Wall

Fig. 2.3. Angle of deviation

White light R

V

δr δv

27

emergent ray is called the angle of deviation. The emergent ray is benttowards the base of the prism.

For a given prism, the angle of deviation of each colour depends on(i) its speed in the prism(ii) the refractive index, µµµµµ of the prism and(iii) its wavelength, λλλλλ

Since the speed, refractive index and wavelength are different for differentcolours, each colour is deviated through a different angle. Of all the colours, violetis deviated the most and red the least.

Recombination of white light

Τwo identical prisms of the same material are arranged side by side such thatone is upright and the other is inverted as shown in Fig. 2.4.

A narrow beam of white light is made to fall on the first prism.The emergent coloured rays are made incident on the second prism. Colouredrays on passing through the second prism emerge as white light.

Since the downward deviation of the colours by the first prism is equal andopposite to the upward deviation by the inverted prism, recombination of colourstakes place. This not only shows that white light is made up of seven colours butalso that the colours are not contributed by the prism which separates them.

Fig. 2.4. Recombination of colours into white light

White lightWhite light

R

V

R

V

R

V

28

Pure and impure spectrum

In the spectrum obtained from aprism, there is overlapping of coloursas each colour merges into the other asshown in Fig. 2.5. A spectrum in whichthere is overlapping of colours iscalled an impure spectrum.

A pure spectrum is one in whichthe various colours obtained on thescreen are distinctly separated fromeach other so that there is nooverlapping of colours.

Conditions for producing a pure spectrum

(i) The beam of light to be dispersed should be from a narrow slit.(ii) The beam incident on the prism should be a parallel beam.(iii) The prism must be in the position of minimum deviation.(iv) The emergent beam should be brought to focus using a converging lens.

Experiment to produce a pure spectrum

White light from a source is made to pass through a very narrow slit S.A convex lens L

1 is placed in front of the slit such that the slit is at its

principal focus. The rays are rendered parallel by the lens and are made to

fall on the refracting face AB of the prism ABC, kept in the positionof minimum deviation. The rays are dispersed by the prism and each colouremerges as a parallel beam as shown in Fig. 2.6.

screen

V

R

A

L2R

V

CV

R

Fig . 2.6. Experiment to produce a pure spectrumB

L1

White light

Fig. 2.5. Impure spectrum

A

B c

R1

R2

V2

V1

Overlappingofcolours

Screen

29

A convex lens L2 is placed on the other side of the prism and a screen is placed

at the focus of the lens L2. A pure spectrum is obtained on the screen.

2.3 Primary and secondary colours

It is found that by projecting any two of red, green and blue light in varyingproportions on a white screen, almost all colours of light can be obtained, but red,green and blue colours cannot be obtained by mixing two other colours. Primarycolours are those which cannot be obtained by mixing other colours.Red, green and blue are primary colours. Secondary colours are thoseobtained by adding any two of the three primary colours.

Complementary colours

A primary colour and the secondary colour obtained by mixing theother two primary colours together produce white. Two colours (a primaryand a secondary colour), which produce white light on mixing are calledcomplementary colours.

Blue and yellow on mixing give white. This is the reason why laundry blueis added, to enhance the whiteness of clothes which have become yellowafter long use.

Red + Cyan → WhiteGreen + Magenta → WhiteBlue + Yellow → White

Fig 2.7. Mixing of colours

R

M Y

GB C

W

RED

BLUE GREEN

Magenta Yellow

White

Cyan

30

Colour mixing

The colours that we see around us are the colours reflected or transmitted bythe bodies. These colours may be spectral colours or combination of coloursobtained by mixing two colours. There are two methods of mixing in whichdifferent colours are obtained. They are by (i) addition and (ii) subtraction.

(i) Addition of coloursThe secondary colours magenta, cyan and yellow are obtained by mixing or

adding two primary colours. This is called colour mixing by addition.

(ii) Subtraction of coloursPaints and dyes are pigments which are colouring matter. Pigment colours are

different from spectral colours. Pigments reflect some colours and absorbthe remaining colours.

When two paints are mixed together more colours are absorbed thanreflected. So the final colour of the mixture is the colour they both reflect. It is notthe same as what is obtained when light beams overlap.

Example: Yellow light + Blue light → White light

Yellow paint + Blue paint → Green paint

This is because pigments in common use are impure colours. Whenilluminated by white light, yellow paint reflects red, yellow and green and absorbsblue. Blue paint reflects blue and green, absorbs yellow and red as illustratedbelow.

The only colour they both reflect is green. Hence, the mixture of yellow andblue pigments looks green. Between them they absorb (subtract) red, yellow and

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123456789011234567890112345678901123456789011234567890112345678901

123456789012123456789012123456789012123456789012123456789012123456789012

123456789011234567890112345678901123456789011234567890112345678901

RED GREENYELLOW

GREEN BLUE

Absorbed

Reflected

YELLOWRED

BLUEYELLOW

BLUE

31

blue from white light. This process is known as colour mixing by subtraction.Cyan, yellow and magenta form primary pigments.

Colour of objects in white light

The colour of a body is that which it reflects. When white light is incident on thebody, it reflects its own colour and it absorbs all the other colours.

(eg.) A green leaf reflects green and absorbs the other colours.

A red rose reflects red and absorbs all the other colours.

If a body reflects all the colours that fall on it, it appears white.

If a body absorbs all the colours that fall on it, it appears black. Sothe black is the absence of all colours.

Colour filters

Filters are pieces of coloured glass, plastic or sheets of gelatine which allowonly certain colours to pass through them as shown in Fig. 2.8.

A red filter transmits onlyred and absorbs all the othercolours.

A blue filter transmits onlyblue and absorbs all theother colours.

R

O

Y

G

B

I

V

R

O

Y

G

B

I

V

Red Filter

R

Blue Filter

B

32

A green filter transmits onlygreen and absorbs all theother colours.

A yellow filter transmits notonly yellow but also red,orange and green andabsorbs the other colours.

2.4 Infra-red rays

Infra-red rays were detected by William Herschel, the astronomer, in 1800.They are emitted by the sun and all hot bodies. The radiations beyond thered region of the visible spectrum are called infra-red rays. Theirwavelength ranges from 7.8× 10–7 m to 3× 10–5 m.

Properties of infra-red rays

(i) They travel in straight lines with the velocity of light.

(ii) They obey the laws of reflection and refraction.

(iii) They heat up the bodies on which they fall.

(iv) They are absorbed by glass but are not absorbed by rock salt.

(v) They penetrate through mist, fog and haze.

(vi) They are scattered less because of their long wavelength.

Fig. 2.8. White light passing through colour filters

R

O

Y

G

B

I

V

R

O

Y

G

B

I

V

Green Filter

G

Yellow Filter

GYOR

33

DetectionInfra-red rays are detected using their heating property with the help of (i) a thermopile and (ii) a thermometer having its bulb blackened.

Uses of infra-red rays

(i) Infra-red rays are used in physiotherapy for the treatment of arthritis.

(ii) They are used in long distance, aerial and night photography and tophotograph in mist and fog.

(iii) Infra-red telescope is used for seeing in the dark.

(iv) They are used in the study of structure and nature of chemical bonds ofmaterials.

(v) Infra-red satellite pictures help us to identify water sources on the earth.Water absorbs infra-red radiation. Hence, lakes and river appear dark inthese photographs.

Greenhouse effect and Global warming

Glass allows visible light to pass through but not infra-red rays. During the day,sunlight passes into the greenhouse. Greenhouse is a building constructedmainly of glass to grow and protect plants. During the daytime the sun lightpasses into the green house and is absorbed by the walls, earth, plants and so on.This absorbed visible light is re-radiated as infra-red radiation, which cannotescape the enclosure. The trapped infra-red radiation causes the temperature ofthe interior to rise. As a result, the greenhouse is warm inside.

Carbon-dioxide, methane, nitrous oxide and ozone in the earth’satmosphere act like the glass in the greenhouse. These gases are called greenhouse gases. They allow the visible light from the sun to pass through but not muchof the infra-red rays. The visible light that reaches the earth’s surfaceis absorbed and re-radiated as infra-red radiation . This IR radiation is trappedand absorbed by the earth’s atmosphere. This results in warming up of the earth’satmosphere.

Heating up of the earth’ s atmosphere due to infra-red radiationre-radiated from the earth’ s surface and absorbed by the greenhousegases present in the atmosphere is called greenhouse effect.

34

As fossil fuels are burnt, a large amount of carbon-dioxide is released into theatmosphere, which absorbs Infra-red radiations from the surface of the earth.This raises the temperature of the earth. Consequently the polar ice-caps wouldmelt raising the sea level. This is of great concern to the whole world, as thecoastal areas are likely to be submerged.

Ultra-violet rays

The radiations lying beyond the violet end of the visible spectrum arecalled ultra-violet rays. This part of the spectrum was detected by J. Ritterin 1801. Ultra-violet rays are emitted by electric sparks, carbon arcs, mercuryvapour lamp with a quartz cover and the sun.

Properties

(i) Ultra-violet rays can pass through quartz.

(ii) They travel in straight lines with the velocity of light.

(iii) They obey the laws of reflection and refraction.

(iv) They affect photographic plates.

(v) They cause fluorescence in substances like zinc sulphide, fluorspar,quinine sulphate solution, etc.

Detection

They are detected by their property of causing fluorescence. A thin tubecontaining a solution of quinine sulphate is found to fluoresce with bluish lightwhen placed in the path of ultra-violet rays.

Uses of ultra-violet rays

(i) Ultra-violet rays are used for sterilising surgical instruments.

(ii) They are used in forensic science to detect forgeries, alteration of legaldocuments, adulteration in food and counterfeit currency.

(iii) Vitamin D is produced in plants and humans when exposed to ultra-violetrays in small doses.

(iv) Bottled drinking water is purified and made free from diseaseproducing bacteria by treating it with ultra-violet rays.

35

Depletion of the ozone layer

Ultra-violet rays cause suntan, but over exposure to ultra-violet rays of shortwavelength, causes skin diseases and premature ageing. These rays from the sundo not reach the earth in large amounts as the ozone layer in the stratosphereabsorbs them. But, the ozone layer is being depleted by chlorofluorocarbonsemitted by refrigerants. This is of great concern to mankind.

Know it yourself

Laundry whitening agents contain small quantities of a certainsubstance which fluoresces with a bluish white light when ultra-violet raysfall on them. This enhances the whiteness of clothes.

2.5 Photoelectric effect

When light of suitable frequency is incident on a metal surface,electrons are ejected from the metal. This phenomenon is calledphotoelectric effect. The electrons ejected from the metal are calledphoto electrons.

Experimental arrangement to demonstrate photoelectric effect

Fig 2.9. Experimental arrangement to demonstrate photoelectric effect.

Quartz window

G

Evacuated glasstube

B Rh

P C

36

The apparatus consists of an evacuated glass tube in which aphoto-sensitive plate P is placed at one end and a collecting electrode C atthe other end. A quartz window W is sealed on to the glass tube which permits theultra-violet light to pass through and irradiate the plate P. A battery B andgalvanometer G are connected across the plates P and C as shown inthe Fig. 2.9. The plate C is maintained at a positive potential with respect to P.

Ultra-violet light is allowed to be incident on the plate P. The galvanometershows deflection indicating the flow of current. When the plate C is connectedto the negative terminal of the battery there is no deflection in the galvanometer.

It is clear from the above experiment that when ultra-violet rays are incidenton the plate P, electrons are emitted. These electrons are called photoelectronsand the current is called photoelectric current.

Photoelectric current depends on

(i) intensity of the incident radiation(ii) frequency of the incident radiation and(iii) potential difference between the cathode and the anode.

Einstein’s photoelectric equation

Metals have a large number of free electrons, which wander throughoutthe body of the metal. However, these electrons are not free to leave the surfaceof the metal. A minimum amount of energy is required to liberate an electron fromthe surface of the metal.

According to Einstein, light consists of bundles of energy called photons.The energy of each photon is hυ, where h is Planck’s constant and υ is thefrequency of light. The value of Planck’s constant h = 6.625 x 10–34 Js. When abeam of light is incident on the surface of the metal, each photon of energy hυ

liberates a free electron

The minimum energy required for releasing an electron from thesurface of the metal is hυυυυυ

00000,,,,, which is equal to the work function of the

metal. The frequency υυυυυ0 0 0 0 0 is the threshold frequency.

37

The remaining energy (h υ – h υo) appears as kinetic energy of the electron. If

the electron does not lose energy due to internal collision, then

m v2

max= h υ – h υ

o

This equation is known as Einstein’s photo electric equation.

Laws of photo electric emission

1. For light of any given frequency, the photoelectric current is directlyproportional to the intensity of light provided the frequency is above thethreshold frequency.

2. For a given photosensitive material, there is a certain minimumfrequency called threshold frequency below which the emission ofphotoelectrons stops completely, no matter, how great the intensity oflight may be.

3. Photoelectric emission is an instantaneous process. As soon as thefrequency of light exceeds the threshold frequency, the emission startsimmediately without any apparent time lag.

4. The maximum kinetic energy of the photo electrons is directlyproportional to the frequency of light but is independent of itsintensity.

2.6 Dual nature of lightThe most difficult question to answer regarding light has proved to be,

whether it light consists of a stream of particles or waves. Thecharacteristics of particles are that at any given instant they have energyand momentum, while the characteristics of waves are wavelength andfrequency.

During interference, diffraction and polarisation, light behaves like a wave.During the other phenomena of absorption, emission of light andphotoelectric effect, light behaves like a stream of particles. Hence, lighthas a dual nature.

12

38

2.6.1. de Broglie’s matter waves

In 1924, the French scientist Louis de Broglie reasoned the following conceptfrom the fact that nature is symmetrical. According to him, if nature allows lightto behave particle-like and wave-like, matter should also be allowed to behaveparticle-like and wave-like. Particle in motion is associated with a wave calledmatter wave or de Broglie wave.

Wavelength of matter waves

According to quantum theory, the energy associated with each photon,

E = h υ ……(1)

According to Einstein, when a mass m is completely converted into energy,

E = m c2 where c is the velocity of light.……(2)

From (1) and (2),

h υ = m c2

= m c2 since υ =

λ = ……(3)

For a particle with a velocity v, the equation (3) becomes

λ =

λ = ( ... p = mv, the momentum of the particle)

Here λ is called de- Broglie wave length.

hm c

h c λ

hp

cλ

( )

hm v

39

2.7 Scattering of light

Tyndall was the first to study experimentally the scattering of light. When aparallel beam of light passes through a gas or liquid, a part of it appears indirections other than that of the incident light. This phenomenon iscalled scattering of light. The scattering is due to absorption of light by themolecules of the medium and its re-radiation in different directions. Scatteringof light is of two types

(i) Coherent scattering If the scattered light has the same wavelength asthe incident light it is called coherent scattering. Example: Rayleigh’s scattering.

(ii) Incoherent scattering If the scattered light has a wavelength differentfrom that of the incident light, it is called incoherent scattering. Example: Ramanscattering.

Rayleigh’s scattering Rayleigh studied the scattering of light by airmolecules. He found that the amount of scattering depends on the wavelength ofthe light and also the size of the particle which causes scattering.

Rayleigh’s scattering law states that the amount of scattering of light isinversely proportional to the fourth power of its wavelength.

A ∝1

λ 4

Problem

Calculate de - Broglie wave length of an electron moving with a velocity of106 ms–1.

velocity v = 106 ms–1

de - Broglie wave length = 6.625 x 10–34

λ = 7.3 x 10–10 m

hm v

λ = 9.1 x 10–31 x 106

40

Problem

Compare the amounts of scattering of a beam of violet light of wavelength3800 A and a beam of red light of wavelength 7600 A when they pass through thesame medium.

Amount of violet light scattered = A1

Amount of red light scattered = A2

Wavelength of violet light = λ1

Wavelength of red light = λ2

According to Rayleigh’s scattering law,

=

=

= = 24 = 16

Scattering of violet light is 16 times that of red.

Colour of the skyThe sky appears blue due to scattering of sunlight by the air molecules in the

atmosphere. According to Rayleigh’s scattering law, light of shorter wave lengthscatters more. Blue light being of shorter wave length is scattered to a greaterextent than red light. This causes the sky to appear blue.

At sunrise and sunset, rays of the sun have to travel a larger part of theatmosphere than at noon. Therefore, most of the blue light is scattered away andonly the red light, which is least scattered reaches the observer. Hence, the sunappears reddish at sunrise and sunset.

Red having the longest wavelength of visible light is scattered the least. It cantravel longer distances than other colours without appreciable loss of intensitydue to scattering. It can be seen as a bright red light at a much longer distancethan the other colours. This is why red is used to indicate danger andwarning.

o o

A1

A2

1/λ1

4

1/λ2

4

λ2

4

λ1

4

(7600)4

(3800)4

41

2.8 Raman effect

A new type of molecular scattering entirely

different from Rayleigh’s scattering was

discovered by Sir C.V. Raman in 1928. When

an intense beam of monochromatic light is

passed through a substance, the light is

scattered and the scattered light contains other

frequencies in addition to that of the incident

light. This is known as Raman effect. For the

above discovery Sir C.V. Raman was awarded

Nobel Prize on 10th December 1930.

When the scattered radiation is examined by a spectroscope, a number of newlines are observed on both sides of the parent line corresponding to incidentfrequency. The new lines are called Raman lines. The lines having lesserfrequencies than that of incident light are called Stokes lines.The lines having higherfrequencies than that of incident light are called Antistokes lines.

The difference between the frequency of any Raman line and the frequency ofthe parent line is known as Raman shift. For a given scattering material, the Ramanshifts are the same, whatever be the frequency of the incident light.

Raman shift is used to analyse the chemical composition of different substancesand also to classify them according to their molecular structure.

wtype

Sir C.V. Raman

Know it yourselfThe colour of the sea The blue colour of the sea is partly due to

reflection from the blue sky and partly by scattering of light by thewater molecules. Near the shore the sea is green because of thesand in suspension which scatters blue and green light and also dueto the reflection of yellow light from the sand.

42

Activity

1. Using the computer, try colour mixing. You will be surprised that you canproduce upto 50,000 colours

2. Take three torch lights. Cover the glass of the first with red, the second withblue and the third with green cellophane paper. Flash the red and blue light atthe same spot on a white screen. What colour do you observe on the overlappedregion of the screen?

3. Similarly. observe the overlapped region due to blue and green light and thendue to red and green light. Finally flash all the three colours on the same spot.What do you observe in each case?.

It is observed that,

Case 1 : Red + Blue → Magenta

Case 2 : Blue + Green → Cyan (peacock blue)

Case 3 : Green + Red → Yellow

Case 4 : Red + Blue + Green → White

Thus magenta, cyan and yellow are secondary colours.

Let us muse upon

F The theoretical prediction of electromagnetic waves was first made byJames Clark Maxwell.

F Hertz was the first scientist to produce the electromagnetic wavesexperimentally.

F The velocity of electromagnetic waves is given by c = υ λ

F The FM radio uses electromagnetic waves of frequency ranging from88 MHz to 108 MHz.

F The AM radio uses electromagnetic waves of frequency ranging from550 kHz to 1000 kHz.

43

F The splitting of white or any composite light into its constituentcolours on passing through a prism is called dispersion.

F The angle between the incident ray and the emergent ray is calledthe angle of deviation.

F The angle of deviation of each colour depends on speed, refractiveindex of the prism and its wavelength.

F Violet is deviated the most and red the least.

F A spectrum in which there is overlapping of colours is called animpure spectrum.

F A spectrum in which there is no overlapping of colours is called apure spectrum.

F Primary colours are those colours which cannot be obtained bymixing any other combination of colours.

F Secondary colours are those colours that are produced by mixing anytwo primary colours.

F Complementary colours are those two colours which when addedtogether gives white.

F When an object reflects all the colours of white light, the objectappears white.

F When an object absorbs all the colours of white light, the objectappears black.

F Heating up of the earth’s atmosphere due to infra-red radiationre-radiated from the earth’s surface and absorbed by the greenhousegases present in the atmosphere is called greenhouse effect.

F When light of suitable frequency is incident on a metal surface,electrons are ejected from the metal. This phenomenon is calledphotoelectric effect. The electrons ejected from the metal are calledphoto electrons.

44

F The minimum energy required to liberate an electron from the surfaceof the metal is called work function. It is equal to h υ

o where h is

Planck’s constant and υo is the minimum frequency ofradiation. This

frequency is known as threshold frequency.

F If scattered light has the same wavelength as the incident light, it iscalled coherent scattering or Rayleigh’s scattering.

F If scattered light has a wavelength different from that of the incidentlight, it is called incoherent scattering.

F Rayleigh’s scattering law states that the amount of scattering of light isinversely proportional to the fourth power of its wavelength.

Self evaluation

2.1. Electromagnetic waves were experimentally produced at first by

(a) Maxwell (b) Hertz (c) Planck (c) Faraday

2.2. The property which is common to X-rays, gamma rays, IR rays, UV raysand visible light is that,

(a) they all have similar wavelengths.(b) they are all damaging to health.

(c) they all travel with the same velocity in vacuum.

(d) they are all affected by electric and magnetic fields.

2.3. A VHF radio station broadcasts at a frequency of 90 MHz. The wavelength of the wave broadcast by the station is,

(a) 0.30 m (b) 3.3 m (c) 27 × 1015m (d) 12 × 1014m

2.4. A copper plate is heated to 1000C. It cools by emitting

(a) α rays (b) IR rays (c) UV rays (d) Gamma rays

2.5 1 A is equal to

(a) 10–7m (b) 10–9m (c) 10–10m (d) 10–12m

2.6 The wavelength 5 x 10–7m corresponds to

(a) IR spectrum (b) X rays spectrum (c) visible spectrum (d) UV spectrum

o

45

2.7. The wave length range of visible spectrum is

(a) 3.8 x 10–7m to 7.8 x 10–7m (b) 7.8 x 10–7m to 3.8 x 10–5m(c) 6 x 10–10m to 4 x 10–7m (d) 1 x 10–10m to 3 x 10–8m

2.8. Which of the following is correct?

(a) red + green → cyan (b) red + blue → magenta

(c) green + blue → white (d) red + blue → green

2.9. The ratio of the wavelength of two colours is 2:3. The ratio of the amount ofthese colours scattered by air molecules is

(a) 2:3 (b) 3:2 (c) 24 : 34 (d) 34 : 24

2.10. Blue of the sky is due to

a) dispersion of light (b) refraction of light

(c) deviation of light (d) scattering of light

2.11. The particles emitted during photoelectric emission are

(a) α particles (b) electrons

(c) protons (d) neutrons

2.12.The number of photoelectrons emitted during photoelectric emission depends on the

(a) frequency of the incident light (b) intensity of the incident light

(c) potential difference between cathode and anode

(d) all the above

2.13. The maximum kinetic energy of photoelectrons depends on

(a) frequency of light (b) intensity of light

(c) applied voltage (d) all the above

2.14. Which colour deviates the most on passing through a prism?

(a) Red (b) Blue (c) Green (d) Orange

2.15. The complementary colour of yellow light is

(a) red (b) blue (c) green (d) orange

2.16. What is an electromagnetic wave?

2.17. Mention any two properties of electromagnetic waves.

46

2.18. Give the uses of microwaves.

2.19. How are radio waves produced? Mention their applications.

2.20. What is dispersion?

2.21. Define angle of deviation. On what factors does it depend?

2.22. What is a pure spectrum? Mention the conditions for obtaining it.

2.23. Explain with a neat diagram, how a pure spectrum is produced.

2.24. What are primary colours?

2.25. What are secondary colours? Give examples.

2.26. What are complementary colours? Give examples.

2.27. Explain the two methods of colour mixing.

2.28. What are primary colours in pigments?

2.29. What are the properties of IR rays?

2.30. What is green house effect?

2.31. Mention the properties of UV rays.

2.32. Give the uses of UV rays.

2.33. What is the cause for the depletion of the ozone layer? How does it affectmankind?

2.34. Define photoelectric effect.

2.35. Describe an experiment to study photoelectric emission.

2.36. Derive Einstein’s photoelectric equation.

2.37. State the laws of photoelectric emission.

2.38. On what factors does the photoelectric current depend?

2.39. Define work function.

47

2.40. Derive an expression for de–Broglie wave length.

2.41. State Rayleigh’s scattering law.

2.42. Why does the sky appear blue?

2.43. The sun appears reddish during sun set or sun rise. Why?

2.44. Write a short note on Raman effect.

2.45. Signals warning danger use red light. Give reason.

2.46. Calculate the momentum of a particle associated with a wave of

wavelength 2 A .

2.47. The work function of a photosensitive material is 26.5 x 10–17 joule.Calculate the threshold frequency.

2.48. What is the energy of a photon of frequency 7.5 x 1014 Hz ?

2.49. Calculate the wavelength associated with a particle of mass 5 x 10–24 kg moving with a velocity 2 x 107 ms–1.

2.50. When a mass of one gram is completely converted into energy how much energy is released?

Answers

2.1. (b) 2.2. (c) 2.3. (b) 2.4. (b) 2.5. (c)

2.6. (c) 2.7. (a) 2.8.(b) 2.9.(d) 2.10.(d)

2.11. (b) 2.12. (b) 2.13. (a) 2.14.(b) 2.15. (b)

2.46. 3.313 x 10–24 kg ms–1 2.47. 4 x 1017 Hz

2.48. 4.969 x 10–19J 2.49. 6.625 x 10–18 m

2.50. 9 x 1013 Joule

°

48

3. ELECTRICITY

In 1820, the Danish scientist Hans Christian Oersted discovered that a currentcarrying conductor has a magnetic field associated with it andestablished the magnetic effect of electric current. This discovery providedthe first link between electricity and magnetism. In 1821, the French scientistAmpere showed that a current carrying conductor placed in a magneticfield experiences a mechanical force. This is called mechanical effect ofelectric current.

3.1 Mechanical effect of current

The mechanical effect of an electric current can be demonstrated by thefollowing experiment.

A light metal roller CD is placed on two metallic rails. This is placedbetween the poles of a strong magnet which is kept along the magnetic meridian

Fig. 3.1. Mechanical effect of electric current

N

N

D

A

B

C

S

E

as shown in fig. 3.1. The ends of the metallic rails are connected to a battery,tap key and an ammeter in series. When the key is pressed the current flowsalong BCDA and the roller moves towards the West. When the direction offlow of current is reversed, the roller moves towards the East.

49

The following inferences are made from this experiment.

(i) The current carrying conductor placed in a magnetic field experiences amechanical force.

(ii) The direction of the force is perpendicular to the direction of both themagnetic field and the current flowing through the conductor.

Fleming’s left hand rule (Motor rule)

Since this rule is applied

in the electric motor, it is

also called the motor rule.

Fleming’s left hand rule is

used to find the direction of

the mechanical force on a

current carrying conductor

placed in a magnetic field.

Stretch the first three fingers of the left hand mutually at right angles toeach other. If the forefinger represents the direction of the magnetic fieldand the middle finger represents the direction of the current, then thethumb represents the direction of the force.

Magnitude of the mechanical force The magnitude of the mechanical force ‘F’ acting on a current carrying

conductor held perpendicular to the direction of the magnetic field isproportional to the current ‘I’ and the length ‘l’ of the conductor.

i.e. F ∝ I

F ∝ l

∴ F ∝ I l

F ===== B I lllll

where B is the magnetic field or magnetic induction.

Fig. 3.2. Fleming’s left hand rule

N S

current

field

motion

F

I

50

The force is maximum when the conductor is perpendicular to the field and iszero when it is parallel to the field.

ProblemA straight wire of length 0.2 m stretched horizontally carries an electric

current of 10 A from East to West in a magnetic field of induction 10–3 tesla,directed vertically downwards. Find the magnitude and direction of the force.

l = 0.2 m

I = 10 A

B = 10–3 T

F = B I l

= 10–3 × 10 × 0.2 N

force F = 2 × 10–3 N and it acts towards the South.

3.2 Loudspeaker

Loudspeaker converts electrical energy into sound energy. It works on theprinciple that a current carrying conductor placed in a magnetic fieldexperiences a force.

Fig. 3.3 Loudspeaker

N

S

N

Electric signal from the amplifier

Voice CoilSpecially treated paper

The loudspeaker consists of a specially shaped permanent magnet. A thincylindrical insulated voice coil is placed such that it is free to move in the radial

51

field of the magnet as shown in Fig. 3.3. The coil is attached to a specially treatedpaper in the form of cone.

The amplified electric signals are passed through the voice coil. The coilexperiences a force which makes it move to and fro. The motion of the coil causesthe cone to vibrate, thus causing the surrounding air molecules to vibrate. Due tothis vibration, sound is reproduced with the original frequency.

3.3 Galvanometer

Galvanometer is an instrument for detecting the current flowing inan electric circuit. It is based on the principle that a current carryingconductor placed in a magnetic field experiences a force. There are two types ofgalvanometer.

1. Moving coil galvanometer.2. Moving magnet galvanometer.

Moving coil galvanometer is further classified into suspended coilgalvanometer and pivoted coil galvanometer.

3.3.1 Pivoted moving coil galvanometer

The essential parts of the pivoted type moving coil galvanometer are

(i) a powerful horse-shoe magnet

(ii) insulated rectangular coil

(iii) cylindrical soft - iron core

(iv) springs and

(v) pointer and scale

It consists of a rectangular coil ABCD of a large number of turns ofinsulated copper wire wound on a frame of a non-magnetic material likealuminium. A soft-iron cylinder is placed inside the coil. This coil is pivotedbetween two bearings X and Y and placed between two concave poles N and Sof a permanent horse-shoe magnet. The coil is free to rotate in thecylindrical gap between the poles of the magnet. There are twophosphor-bronze hair springs S

1 and S

2 to provide a restoring couple.

52

Fig. 3.4. Moving coil galvonometer

A pointer P is attached to the coil and can move on a calibratedscale. The current to be detected is passed through terminals T

1 and T

2 and it

flows through the coil.

When the current passes through the coil, the vertical sides AB and CDexperience equal and opposite force, according to Fleming’s left hand rule.These forces constitute a deflecting couple. This deflecting couple rotates the coiluntil it is balanced by the restoring couple provided by the springs S

1 and S

2.

The deflection of the coil is indicated by the pointer on the graduated scale.This deflection is proportional to the current.

The galvanometer is converted into an ammeter by connecting a lowresistance in parallel with it. This low resistance is called as shunt.The galvanometer is converted into a voltmeter by connecting a high resistance inseries with it.

25

20

15 10

5 0 5 10

15

20

25X P

A

scale

S

rectangularcoil

soft-iron core

T1

BC

Y

T2S2

S1

D

F

FN I

I

53

3.4 Electric motor

An electric motor is a device which converts electrical energy intomechanical energy. The principle of the electric motor is based on the mechanicaleffect of current, i.e, a current carrying conductor placed in a magnetic fieldexperiences a mechanical force.

D C Motor

The essential parts of a DC motor are

(i) Field magnet The field magnet is a powerful magnet.

(ii) Armatur e The armature ABCD is a rectangular coil of wire wound on asoft-iron core and placed between the poles of the magnet. It is capable ofrotating about its axis.

(iii) Split rings (Commutator) The split rings also known as commutator ismade of copper. The ends of the coil are connected to the split ringsS

1 and S

2 which are capable of rotating with the coil.

(iv) Carbon brushes Two carbon brushes B1 and B

2 are in contact with the

split rings. They are connected to the battery.

Fig. 3.5. D C motor

N

DA

B C

S

B1

B2

S2

S1

54

When the circuit is closed, current flows in the arms AB and CD of the coilin opposite direction. According to Fleming’s left hand rule AB experiences adownward force and CD experiences an upward force, equal in magnitude andopposite in direction. These two forces constitute a couple which rotates the coilin the anticlockwise direction till S

1 and S

2 come in contact with B

2 and B

1

respectively. Now the direction of the current and hence the direction of force onAB and CD is reversed. AB experiences an upward force and CD a downwardforce. Again the couple rotates the coil in the anticlockwise direction. Thus the coilcontinues to rotate in the anticlockwise direction as long as the current flows in it.

The power of a DC motor can be increased by,

(i) increasing the number of turns in the coil. (ii) increasing the strength of current and

(iii) increasing the strength of the magnetic field.

3.5 Magnetic flux and magnetic induction

Magnetic flux

The total number of magnetic lines of force crossing a given area isknown as the magnetic flux. It is denoted by φφφφφ.....

Magnetic field in a region ischaracterised by magnetic lines of force.

The SI unit of magnetic flux is weber(Wb). It is a scalar quantity.

Magnetic induction

The total number of magnetic linesof force crossing per unit area heldnormal to the surface is calledmagnetic induction. It is denoted by B

(i.e.) φ

The SI unit of magnetic induction is tesla (T). It is a vector quantity.

1 tesla = 1 Wb m–2

Fig. 3.6. Magnetic flux

B = A

55

Problem

Calculate the magnetic flux linked with a coil of area 0.5 cm2 placed at rightangle to a magnetic field of induction 0.1 tesla.

A = 0.5 cm2

= 0.5 × 10–4 m2

B = 0.1 tesla

Magnetic flux φ = B A

= 0.1 × 0.5 × 10–4 Wb

Magnetic flux φ = 5 × × × × × 10–6 weber

3.6 Electromagnetic induction

After the discovery of the magnetic effect of current in 1831, Michael Faraday

succeeded in proving the reverse effect. He produced induced emf by the effect of

a varying magnetic field.

Faraday’s experiments

A coil is connected to a sensitive

galvanometer. There is a deflection in the

galvanometer when a magnet is moved towards

the coil. When the magnet is moved away from

the coil, the deflection is in the opposite direction.

The same effect is observed when the magnet

is kept stationary and the coil is moved towards

or away from the magnet. The current obtained

due to the relative motion between the coil and

the magnet is called induced current and the

corresponding emf (electromotive force) is known

as induced emf. Fig. 3.7.

N

N

56

When the magnetic flux linked with a coil changes, an emf is induced init. This phenomenon is called electromagnetic induction.

The induced emf depends on the(i) number of turns of the coil.(ii) magnetic induction and(iii) relative speed between the magnet and the coil.

3.7 Laws of electromagnetic induction

Faraday’s Laws

1. Whenever there is a change in the magnetic flux linked with a coil an emf isinduced in it.

2. The magnitude of the induced emf is equal to the rate of change of flux linkedwith the coil.

If the flux changes from φ1 to φ

2 in time interval t, then

induced emf e =

Lenz’s law

Lenz’s law states that the induced current flows in such a direction that itopposes the change or cause that produces it.

e = –

If there are N turns in the coil, then the induced emf

e = – N

Lenz’s law - In accordance with the law of conservation of energy.

As the magnet is brought towards the coil, the number of magnetic lines offorce linked with the coil changes producing an induced emf. According toLenz’s law the direction of the induced emf is to oppose the motion of themagnet. Hence work has to be done against it to move the magnet further.

( )φ2 – φ

1

t

( )φ2 – φ

1

t

( )φ2 – φ

1

t

57

This work is converted into electrical energy. On the contrary, if the direction ofthe induced emf were to help the motion of the magnet, the magnet will begin tomove faster producing electrical energy. i.e. electrical energy is producedwithout any external work being done, which is impossible according tothe law of conservation of energy. Therefore the induced current acts in adirection to oppose the motion of the magnet.

Problem

Find the emf induced in a coil of 100 turns, if the flux linked with the coil changesfrom 0.5 Wb to 1 Wb in 0.02 s

φ1 = 0.5 Wb, φ2 = 1 Wb, t = 0.02 s N = 100 turns

e = – N

e = – 100

= –

e = – 2.5 × 103 V

Here, negative sign signifies Lenz’s law

Fig 3.8.

100 × 0.50.02

( )1 – 0.50.02

( )φ2 – φ

1

t

NS

58

3.8 Fleming’s right hand rule

Stretch the first three fingers ofthe right hand mutually at rightangles to each other. If theforefinger represents thedirection of the magnetic fieldand the thumb represents thedirection of the motion of theconductor, then the middlefinger represents the directionof the induced current. This ruleis also known as the generatorrule.

3.9 Applications of electromagnetic induction

3.9.1 Microphone

The microphone works on the

principle of electromagnetic

induction. It converts sound wave

into electrical signals which vary in

amplitude and frequency in

accordance with the sound wave.

The microphone has a

diaphragm with a thin coil of wire

attached to it as shown in Fig. 3.10.

The coil is placed in a strong radial

field of a cylindrical pot magnet. Sound waves cause the diaphragm and the coil

to vibrate. As the coil vibrates in the magnetic field, a varying emf is induced in it.

This emf is amplified and then connected to the loudspeaker. Thus sound

variations are converted into electrical variations with the help of a microphone.

Fig. 3.10. The Microphone

PotMagnet

Toamplifier

N

S

N

Coil

Diaphragm

Fig. 3.9. Fleming’s right hand rule

NS

current

field

motion

F

I

59

3.9.2 Generator

A generator is a device used to convert mechanical energy into electricalenergy. There are two types of generator - A C generator and DC generator.

A C generator

The A C generator works onthe principle of electromagneticinduction. The essential parts ofA C generator are

(i) Field magnet It is apowerful horse-shoe magnet.

(ii) Armatur e It is a rectangular coil ABCD of many turns wound on asoft-iron core. It is placed between the poles of the powerful magnet. Thearmature is capable of being rotated about its axis perpendicular to themagnetic field.

(iii) Slip rings The ends of the coil are connected to two hollow metalliccylinders called slip rings S

1 and S

2.

(iv) Carbon brushes B1 and B

2 are two carbon brushes which are

always in contact with the slip rings.

Fig. 3.12. Alternating emf

NDA

B C

S

Fig. 3.11. A C generator

B1

B2

S2

S1

5T/2T/2

T

3T/2

2T

t

R

When the coil is rotated about an axis perpendicular to the direction ofthe magnetic field, there is a continuous change in the magnetic flux linked with it.

em

f

60

An emf is induced in the armature. The emf varies from zero to maximum andthen from maximum to zero for every half rotation of the coil. During the secondhalf rotation the direction of the induced emf is reversed as shown in Fig. 3.12.

The cycle is repeated as long as the armature is rotated. The direction of theinduced emf at any instant is given by Fleming’s right hand rule.

3.9.3 D C generator

The DC generator converts mechanical energy into electrical energy. It workson the principle of electromagnetic induction. It is used to deliver a unidirectionalcurrent.

The essential parts of the DC generator are

(i) Field magnet It is a powerful horse-shoe magnet.

(ii) Armatur e It is a rectangular coil ABCD of many turns wound on asoft-iron core. It is placed between the poles of the horse-shoe magnetand is capable of being rotated about an axis perpendicular tothe magnetic field.

Fig. 3.13. D C generator

(iii) Split ring or commutator The two ends of the coil are connected to thesplit rings S

1 and S

2. They are used to reverse the current.

N

DA

B C

S

B1

B2

S2

S1

R

61

(iv) Carbon brushes Carbon brushes B1 and B

2 make contact with the

split rings S1 and S

2. The induced current passes to the external circuit through

B1 and B

2.

During the first half rotation of the armature, the induced current increasesto a maximum from zero and then decreases to zero. If the armature is rotatedin the anti-clockwise direction, B

2 becomes negative and B

1 positive. During the

second half rotation, the split rings change their contact with the brushes. Thoughthe current in the coil is reversed B

2 remains negative and B

1 positive. As a

result, unidirectional current flows in the external circuit as shown in Fig. 3.14.

The emf obtained from an A C or DC generator can be increased by

(i) increasing the number of turns of the coil.(ii) using a soft iron core.(iii) increasing the speed of rotation and(iv) using a strong field magnet.

Comparison of A C and DC

Alternating current changes its direction periodically and it canbe transmitted over a long distance at a low cost, while direct currentis unidirectional and transmitting it over a long distance is difficult andexpensive.

Fig. 3.14. Unidirectional current

curr

ent

T/2 T 2T3T/2

t

5T/2

Unidirectional current

62

3.9.4 Transformer

A transformer is an electric device used to transfer electrical energyfr om one circuit to another. It is used to step up or step down AC voltage.It works on the principle of electromagnetic induction.

The transformer consists of insulated copper coils wound on either side of alaminated soft-iron core as shown in Fig.3.15.

The input A C voltage is applied to the primary coil P and the output is drawnfrom the secondary coil S.

The alternating current in the primary coil produces a varying magnetic flux.This magnetic flux links not only the primary coil but also the secondary coil.According to the phenomenon of electromagnetic induction the varying magneticflux sets up an induced emf in the secondary coil.

Np and N

s are the number of turns, V

p and V

s are the voltages and I

p and I

s are

the current in the primary and secondary coils respectively.

Since the same change of flux is linked with the primary and thesecondary coil, the induced emf in each turn of the primary and secondary coils isthe same.

Fig. 3.15 Transformer

P S

soft-iron core

o

vs

63

(i.e.) =

= ……… (1)

In an ideal transformer, input power is equal to output power.

input voltage × input current = output voltage × output current.

VP × I

P= V

S × I

S

= ……… (2)

From (1) and (2) = = = K

where K is called turns ratio or transformer ratio.

If K > 1, it is a step up transformer. In this case, the voltage is increased and thecurrent is decreased.

If K < 1, it is a step down transformer. In this case the voltage is decreased andthe current is increased.

3.10 Electric power transmission - Role of transformers

Electricity is generated at thermal, hydro and nuclear power stations,which are situated far from the towns. It has to be transmitted anddistributed to the consumers.

The electric power from the generating station is transmitted to the variousconsumer points using transmission lines. As current flows through thetransmission lines, it heats up the conductor and electric energy is dissipated inthe form of heat equal to I2 R t.

Vs

Vp

Ip

Is

Vs

Vp

Ip

Is

Ns

Np

Vp

Np

Vs

Ns

Vs

Vp

Ns

Np

64

In order to minimise this power loss, the electric power should be transmitted athigh voltage and low current. Hence, the voltage is stepped up at the generating

station using a step-up transformer.

The stepped up voltage cannot be supplied to consumers, as all the electric

appliances are designed to operate at a lower voltage of 220V for domestic

appliances and 440 V for industries. Using step - down transformer the voltage is

stepped down, with the corresponding increase in current and power is delivered

to the consumer at the required voltage.

Step down

transformer City Consumer

Point

Stepup

transformerGenerating

Station

Line wire

Fig. 3.16. Electric power transmission

Problem

1. In a step-up transformer the number of turns in the primary and

secondary are in the ratio of 1000 and 2000 respectively. If the

primary current is 10 A, calculate the secondary current.

Np

= 1000

Ns

= 2000

Ip

= 10 A

65

2. A transformer in a distribution station reduces AC voltage from 36,000 V

to 2400 V. The primary coil has 15000 turns. What is the number of turns

in the secondary?

input Voltage Vp

= 36000 V

output Voltage Vs

= 2400 V

number of turns in the primary Np = 15000

=

=

Ns

=

number of turns in the secondary Ns = 1000

Vs

Vp

Ns

Np

15000 × 240036000

Ns

15000240036000

Ns

Np

Ip

Is

=

2000

1000=

10

Is

Is =

10 x 10002000

=

Is = 5 Asecondary current

102

66

3. The primary current in an ideal transformer is 5 A. When the primaryvoltage is 90 V. Calculate the voltage across the secondary when a currentof 0.9 A is drawn from it.

primary current, Ip

= 5 A

primary Voltage, Vp

= 90 V

secondary current, Is

= 0.9 A

Vs

= × Vp

Vs

= × 90

secondary voltage Vs = 500V

4. A transformer in a radio changes the 240 V AC mains voltage to 12V.What is the turns ratio?

3.11 Electric power

The rate at which electrical energy is consumed in an electric circuit isknown as electric power.

The voltage rating and the power rating are specified on every electricalappliance. e.g. An electric bulb is marked as 240 V, 60W.

The SI unit of electric power is watt (W). The power is said to be onewatt if one joule of energy is consumed by an electrical appliance in onesecond.

The other units of power in common use are megawatt (MW) and horse power(H P) where

1 MW = 106 W1 HP = 746 W

Ip

Is

5

0.9

Ns

Np

Vs

Vp

=12

240= = 0.05turns ratio

67

The commercial unit of electric energy is kilowatt hour(kWh).It is also referred to as a unit.

Kilowatt hour is defined as the electrical energy consumed by anelectric circuit at the rate of one kilowatt in one hour.

Energy = Power × time1 kilowatt hour = 1 kilowatt × 1 hour

= 1000 × 60 × 60 J

1 kWh = 3.6 ××××× 106 J

Problem

1. What is the cost of using a 100 W lamp for 6 hours, at the rate ofRs. 2/- per unit.

energy consumed= Power × time

= 100 W × 6 hour

= kW × 6 hour

= 0.6 kWh or unit

cost per unit = Rs.2/-

∴ total cost = Rs. 0.6 × 2

= Rs. 1.20

2. Calculate the cost of energy consumed for the period from May 1toJune 30 at the rate of Rs.2/- per unit, from the data given below.

Appliances Number duration /day (hour)

60 W lamp 6 4

100 W lamp 4 3

750 W electric iron 1 2

1000 W water heater 1 2

60 W fan 2 10

1001000

68

total energy consumed per day

= (6 × 60 × 4) + (4 × 100 × 3) + (1 × 750 × 2) + (1 × 1000 × 2) + (2 × 60 × 10)

= 1440 + 1200 + 1500 + 2000 + 1200

= 7340 Wh

= 7.34 kWh

total number of days = 31 + 30 = 61

total energy consumed= 61 × 7.34 unit

= 447.74 unit

cost per unit = Rs.2/-

∴ total cost = Rs. 447.74 × 2

= Rs. 895.48

Activity

1. Fill in the blank boxes below

Electric kettle 3000

Tube light 50 10

Vacuum cleaner 500

Washing machine 6 3

Electric iron 750 1

Immersion heater 1200 2

2. Draw a block diagram to indicate the electric connection from the poleoutside your house to the lamp on your study table.

Appliance Power(W)

Power(kW)

Duration ofuse (hr).

Energyconsumed

(kWh)

12

12

34

69

3. Find the importance of earthing and neutral wire for electrical connections.

4. Draw a block diagram to indicate the public address system for thetransmission of sound in your school.

Let us muse upon

F Hans Christian Oersted discovered the magnetic effect of electric current.

F Fleming’s left hand rule is used to find the direction of the mechanical forceon a current carrying conductor placed in a magnetic field.

F The force on a current carrying conductor is maximum when it is heldperpendicular to the magnetic field and zero when it is parallel to thefield.

F Loud speaker converts electrical energy into sound energy.

F Loudspeaker works on the principle of mechanical effect of current.

F The galvanometer is an instrument used for detecting current flowing inan electric circuit.

F The galvanometer is converted into an ammeter by connecting a lowresistance in parallel with it.

F The galvanometer is converted into a voltmeter by connecting a highresistance in series with it.

F Electric motor is a device which converts electrical energy intomechanical energy.

F The total number of magnetic lines of force crossing a given area isknown as the magnetic flux.

F The total number of magnetic lines of force crossing unit area heldnormal to the magnetic lines is called magnetic induction.

F Michael Faraday discovered the phenomenon of electromagneticinduction

F When the magnetic flux linked with a coil changes, an emf is inducedin it. This phenomenon is called electromagnetic induction.

70

F The magnitude of the induced emf is equal to the rate of change offlux linked with the coil.

F The induced current flows in such a direction that it opposes thechange or cause that produces it.

F The direction of induced emf at any instant is given by Fleming’s righthand rule.

F Microphone converts sound energy into electrical energy.

F Generator is a device used to convert mechanical energy intoelectrical energy

F Transformer is an electric device used to transfer electrical energy fromone circuit to another.

F In an ideal transformer, input power is equal to output power.

F The ratio of number of turns in secondary coil to the number of turnsin the primary coil is called turns ratio.

F The rate at which electrical energy is consumed in an electric circuitis known as electric power.

F The power is said to be one watt if one joule of energy is consumed byan electrical appliance in one second.

Self evaluation

3.1. Electrical energy is converted into mechanical energy in a

(a) dynamo (b) transformer (c) motor (d) microphone

3.2. A galvanometer is converted into an ammeter by connecting

(a) a low resistance in series. (b) a low resistance in parallel(c) a high resistance in series (d) a high resistance in parallel

3.3. The output and input voltages in a transformer are 25V and 5V respectively.The transformer ratio is

(a) 30 (b) 125 (c) 5 (d) 1/5

71

3.4. A transformer is used to convert 240V into 12V . If there are 2000 turns in theprimary coil, the number of turns in the secondary coil is,(a) 100 (b) 120 (c) 200 (d) 24000

3.5. Which of the following is based on electromagnetic induction ?(a) transformer (b) galvanometer(c) loudspeaker (d) motor

3.6. The working of the galvanometer is based on(a) magnetic effect of electric current(b) mechanical effect of electric current(c) electromagnetic induction(d) heating effect of electric current

3.7. The direction of induced current is given by(a) Lenz’s law (b) Faraday’s law(c) Fleming’s left hand rule (d) None of the above

3.8. Explain the function of a loudspeaker with a neat diagram.

3.9. State Faraday’s laws of electromagnetic induction.

3.10.Draw the diagram of a galvanometer and explain its construction andworking

3.11. Explain the construction and working of a transformer.

3.12. Explain the role of transformers in the transmission of power.

3.13. State Lenz’s law and explain how it is in accordance with the law ofconservation of energy.

3.14. Describe an experiment to show the mechanical effect of current.

3.15. State Fleming’s left hand rule.

3.16. On what factors does the magnitude of the force on a current carryingconductor in a magnetic field depend?

3.17. Name the two types of galvanometer.

3.18. How will you convert a galvanometer into an ammeter?

72

3.19. How will you convert a galvanometer into a voltmeter?

3.20. Describe the construction and working of an electric motor.

3.21. Define magnetic flux. Give its unit.

3.22. Define magnetic induction. Give its unit.

3.23. Explain Faraday’s experiment on electromagnetic induction.

3.24. State Fleming’s right hand rule.

3.25. Write a short note on microphone.

3.26. What are the essential parts of an AC generator?

3.27. Describe the construction and working of an AC generator.

3.28. Write a short note on DC generator.

3.29. Distinguish between alternating current and direct current.

3.30. Define turns ratio.

3.31. Define electric power.

3.32. Define one kilowatt hour

3.33. 100kW of power is being supplied to a factory through wires of resistance0.1Ω. Calculate the power loss in the wires if the power is transmitted at(a) 230V (b) 10000V.

3.34. Find the number of turns in the secondary coil of a transformer having 3000turns in the primary coil. It operates from a 240V A C supply to give anoutput of 8V for a door bell.

3.35. An immersion heater works on 230 V and draws 4 A of current. If the costof 1kWh is Rs.2/- find the cost of running it for 15 minutes every day for30 days.

3.36. In a house two electric lamps each and of 40 W is used for 6 hours, three fansof 60W each is used for 7 hours and an electric oven of 800 W is used for2 hours per day. Calculate the cost of energy consumed for 30 days if therate per unit is Rs. 2.

3.37. A conductor of length 25 cm is placed perpendicular to a magnetic field ofinduction 0.5 x 10–3 tesla. Calculate the force on the conductor, if a current of1.5 A flows through the conductor.

73

3.38. The magnetic flux linked with a coil changes from 0.3 Wb to zero in1.2 second. Calculate the induced emf.

Answers

3.1. (c) 3.2. (b) 3.3. (c) 3.4. (a)

3.5. (a) 3.6. (b) 3.7. (a)

3.33.(18.9 kw, 10 w ) 3.34. (100)

3.35.(Rs. 13.80) 3.36. (Rs. 200.40)

3.37. (1.87 x 10–4 N) 3.38. (0.25 V)

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4. PROPERTIES OF MATTER

Matter is made up of molecules and atoms. It exists in three statesnamely solid, liquid and gas. The three states of matter can be explained onthe basis of interatomic or intermolecular forces. The force of attractionbetween any two molecules is called intermolecular force. The three states ofmatter differ from each other because of the difference in the magnitude ofthe intermolecular forces. The interatomic force is so strong in solids that itsatom cannot move freely. That is why, the solid has a definite shape and size.In the case of liquids, the intermolecular force is relatively small and hencethe molecules are free to move inside the whole volume of the liquid. Theliquid has a definite volume, but no definite shape. Intermolecular force isvery weak in gases and hence, the molecules of a gas may be considered asfree particles. They keep on colliding against each other and move in a

random manner. A gas has neither a definite shape nor a definite volume.

4.1 Solids

At ordinary temperature, solids possess a definite shape and volume.The properties of the solid depend on the arrangement of atoms in it. Basedon their internal structure, solids are classified into two types – crystalline andamorphous.

4.1.1 Crystalline solid

In crystalline solids, the atoms are arranged in a regular, repeated andperiodic pattern. This orderly arrangement of atoms resembles that of bricklaying by masons. Examples of crystalline solids are diamond, quartz,rocksalt, mica, sugar, metals, etc.

4.1.2 Amorphous solid

In an amorphous solid, the atoms are arranged in a disorderly manner.The best example for an amorphous solid is glass. Other examples are plasticmaterials, wood, etc.

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4.1.3 Elasticity

When an external force ‘F’ of sufficient magnitude is applied on a bodyof mass ‘m’, the body acquires an acceleration ‘a’ such that a = F/ m(Newton’s II law of motion)

However, if the body is not free to move, the external force acting onthe body will change the relative position of the molecules. Due to thischange, the body may suffer a deformation. This external force is known asdeforming force.

As the body is deformed, internal forces are set up within the body,which tends to bring the body back to the original shape. The forcedeveloped within the body on account of relative molecular displacement iscalled internal force or elastic force or restoring force. Under the influence ofthis restoring force, the body regains its original shape after the deformingforce is removed. Such bodies are called elastic bodies. Elasticity is theproperty of the material of a body by virtue of which the bodyregains its original shape when the deforming force is removed.

There are a few bodies, which do not show any tendency to recovertheir original shape after the removal of deforming force. Such bodies arecalled plastic bodies. The property by virtue of which the body does notregain its original shape after the removal of the deforming force iscalled plasticity.

Restoring force developed in an elastic body during deformation isquantitatively equal to the external force applied to cause the deformation.The stress is defined as the restoring force developed per unit area ofthe body. Its unit is N m-2

If there is a change in the shape of a body by the application of a deformingforce, then the body is said to be strained. The strain produced in a bodyis defined as the ratio of change in dimension to its original dimension.

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The maximum value of stress within which a body regains itsoriginal state is called elastic limit.

Hooke’s law

This law was proposed by Robert Hooke, the founder of the RoyalSociety in 1676. He showed that the extension of an elastic body is directlyproportional to the force that produces it, provided the extension is withinthe elastic limit.

Hooke’s law states that within the elastic limit of the body, thestress is proportional to the strain produced.

This constant is known as modulus of elasticity.

4.2 Liquids - types of intermolecular force

There are two types of intermolecular forces

1. Force of cohesion It is the force of attraction between the molecules ofthe same substance. The cohesive force is very strong in solids, weak inliquids and extremely weak in gases.

Example We can suspend a heavy load by the thin steel wire because ofthe strong force of cohesion between the atoms of steel wire.

2. Force of adhesion It is the force of attraction between the moleculesof two different substances.

Examples

i. It is because of the force of adhesion, that ink sticks to paper whilewriting.

ii. Strong adhesive force is shown by materials like cement, gum and paintsto walls.

strainstress

= a constant.i.e.

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Water wets glass because the force of cohesion between the water

molecules is less than the force of adhesion between water and glass

molecules. At the same time, mercury does not wet glass because the force

of adhesion between mercury and glass molecules is less than the force of

cohesion between the mercury molecules.

4.2.1 Molecular range and sphere of influence

The molecular range is

defined as the maximum

distance upto which a molecule

can exert an appreciable force

of attraction on another

molecule. It is of the order of

10-9 m for liquids.

The sphere of influence is defined as the imaginary sphere drawn

with a particular molecule as centre and molecular range as radius.

The molecule at the centre exerts a force of attraction on all the molecules

lying within the sphere of influence.

4.2.2 Surface tension

The free surface of a liquid, at rest behaves like a stretched elastic

membrane so as to possess minimum surface area. This behaviour of the

liquid is due to a property called surface tension. The phenomenon of

surface tension can be easily understood from the following illustrations.

Fig. 4.1 Sphere of Influence

10–9 m

molecule

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4.2.3 Illustrations of surface tension of liquids

(i) An insect walks on the surface of

water without getting wet. There is a slight

depression on the surface of water. The force

of surface tension does not act horizontally

but along an inclined direction. The vertical

component of the force of surface tension

balances the weight of the insect as in fig: 4.2.

2. If a greased sewing needle is gently

placed on the surface of water, it floats. The

water surface below the needle gets slightly

depressed. The vertical component of the

force of surface tension balances the weight

of the needle as in fig: 4.3.

3. When a painting brush is dipped in

water, its hairs spread out. If the brush is

taken out, its hairs will cling together as the

free surface of water film between the hairs

contract due to surface tension as in fig: 4.4.

4.2.4 Molecular theory of surface tension

Consider three molecules A, B and C of a liquid in a container.

The molecule A is attracted equally in all directions by the neighbouring

molecules. So the net force acting on this molecule is zero. As the number

of neighbouring molecules in the lower part of the spheres B and C are more

Fig. 4.2 Insect walks on water surface

Fig. 4.3 Needle floats

w = mg

needleT

Fig. 4.4 Spreading outof hairs

79

than the number in the upper part,

there is a net downward force on these

molecules. Here the molecule C

experiences maximum downward

force. This downward force will try

to move the molecules inside the

liquid. If a molecule from the interior

is to be brought to the surface of

the liquid, work must be done against

this downward force. This work done

on the molecule is stored as potential

energy on the surface of the liquid.

For equilibrium, a system mustpossess minimum potential energy.Consequently the free surface tends tohave the least area, decreasing thenumber of molecules on the surfaceand thus making the surface behave

like a stretched elastic membrane.The surface tension is measured as the force acting per unit length. This forceis always tangential to the liquid surface. Thus, surface tension is definedas the force per unit length acting perpendicular to an imaginary linedrawn on the liquid surface, tending to pull the surface apart along theline. Its unit is Nm-1.

4.2.5 Factors affecting surface tension

i. Variation of surface tension by impuritiesThe presence of impurities of any kind on the surface of a liquid

will affect the surface tension. Depending upon the nature of impurities,the surface tension of the liquid either increases or decreases.

Fig. 4.5 Forces in a liquid

Fig. 4.6 Surface tension

A

BC

80

A highly soluble substance like sodium chloride increases thesurface tension. But sparingly soluble substances like phenol or soapreduce the surface tension.

If a piece of camphor is placed on the surface of water, the surface tensiondecreases at the point where it is placed. At other points on the surface ofwater, the surface tension is more. Due to this, there is a forward forcemaking the camphor piece to move. This happens until the entire surfaceof water is contaminated and then the motion of the camphor ceases.

ii. Variation of surface tension with temperatureThe surface tension of a liquid decreases with rise in temperature. There is

a particular temperature at which the surface tension of a liquid becomeszero. This temperature is called critical temperature of that liquid.

4.2.6 Application of surface tension in day-to-day lifei. During a storm, the waves are rough on the sea. The sailors pour tins of

thick oil around their boats or ships. As the surface tension of oil is lesserthan that of water, the addition of oil decreases the surface tension and hencereduces the intensity of the waves.

ii. Mosquitoes breed on the surface of free water. The layer of water dueto surface tension supports them. Oil has lesser surface tension than water.When the oil is sprayed on the surface of water, due to the higher surfacetension of water, the oil is stretched in all directions as a thin film. As thesurface tension of oil is less, the mosquitoes cannot breed on it. Therefore,to prevent the breeding of mosquitoes in rainy season, oil is sprayed onthe surface of water in pools and ponds.

iii. Dirty clothes having grease and oil stains cannot be washed with waterunless some detergent or soap is added. Water does not wet greasy dirt, butwhen a detergent is added, the surface tension of water-oil system is drasticallyreduced. Now water can wet greasy dirt to make washing easier.

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4.3 Capillarity

When a capillary tube is dipped in water, the water rises up in thetube. When the same capillary tube is dipped in mercury, the mercuryis depressed below the free surface of the mercury in the container.The phenomenon of rise or fall of a liquid in a capillary tube is knownas capillarity.

4.3.1 Illustrations of capillarity

i. The oil in a wick rises up through the narrow spaces between the threadsof the wick, which act as fine capillary tubes.

ii. A blotting paper absorbs ink by capillary action. The pores of theblotting paper act as capillaries.

iii. The action of a cotton towel in soaking up the moisture from hands orthe body is due to capillary action of cotton in the towel.

iv. Walls get damped in rainy season due to the absorption of water by thebricks due to capillary action.

v. A pen nib is split at the tip to provide a narrow capillary tube and theink is drawn up to the point continuously due to capillarity.

vi. Roots, trunk, branches and leaves of a tree possess fine capillaries.Water rises upto the top of the tree by capillarity.

Fig. 4.7 Capillarity

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4.4 Viscosity

Solid surfaces in contact exert a frictional force on each other, when onesurface moves on the other. In a similar way, the relative motion of the layersof liquid is restricted by friction. This frictional force between the layers ofthe liquid is known as viscous force. The property of a liquid by virtue ofwhich it opposes the relative motion between its different layers isknown as viscosity of the liquid. Viscosity differs from liquid to liquid. Thegreater the viscosity the less easy it is for the fluid to flow.

4.4.1 Coefficient of viscosityNewton found that the viscous force between the layers of a liquid is(i) directly proportional to the area of the liquid layers in contact(ii) directly proportional to their relative velocity and(iii) inversely proportional to the distance between themIf A is the area of contact between the two layers, v1

and v2 their

respective velocities and ‘x ’ the distance between them, then the viscous force.

F α A

α v 2 – v

1

α 1/x

F α A (v

2 – v

1)

x

F =η A (v

2 – v

1)

--------- (1) xWhere η is a constant known as the coefficient of viscosity.v

2 – v

1 is called the velocity gradient

x

From the equation (1)

If A = 1 square unit and v

2 – v

1 = 1 unit

then F = ηx

83

Thus, coefficient of viscosity of a liquid is numerically equal to theviscous force acting tangentially between two layers of a liquid havingunit area of contact and unit velocity gradient normal to the direction offlow of the liquid. Its unit is Nsm-2

4.4.2 The flow of a fluid through a pipe

The rate at which a fluid flows through a pipe depends on

i. the viscosity of the fluid ii. the dimensions of the tube iii. the pressure difference between the ends

The rate of flow of a fluid is of greater importance in our lives since, itgoverns things like the flow of blood through blood vessels in our body, thetransmission of gas, water or oil through pipe lines for long distances.

4.4.3 Str eamline flowConsider a path of flow of liquid in a pipe along ABC. Let v

1, v

2 and v

3 be

the velocities of the liquid at three different points, A, B and C respectively.

The flow of liquid in the pipe is said to be streamlined if(i) all the particles crossing the point A have the same velocity v

1.

(ii) all the particles crossing the point B have the same velocity v2,

but v2 may or may not be equal to v

1.

(iii) similarly all the particles arriving at the point C have the same velocity v

3.

Fig. 4.9 Streamline flow

AB C

V1V2

V3

84

Thus streamline flow of a liquid is defined as that flow in whichevery particle of the liquid follows exactly the same path of itspreceding particle and has the same velocity in magnitude anddirection as that of its preceding particle while crossing through a point.

4.4.4 Turbulent flow

The flow is streamlined as long as the velocity of the liquid does notexceed a limiting value.When the external pressure causing the flow of liquidincreases, the motion of the liquid takes place with a velocity greater than thelimiting value and the flow of liquid loses all its orderliness. This type of flowis called turbulent flow.

Critical velocity of a liquid is the velocity below which the motionof the liquid is streamlined and above which the motion becomesturbulent.

4.4.5 Reynold number

Reynold number is a mere number, which determines whether the flowof a liquid through a pipe is streamlined or turbulent. Reynold numberK is given by

K = ρvD

η

where η is the coefficient of viscosity of the liquid, ρ is the density of theliquid, v is the average velocity of the liquid through the pipe and D is thediameter of the pipe.

If the value of Reynold number lies between 0 to 2000, the flowof liquid is streamlined. For value of K above 3000, the flow of liquid isturbulent and for values of K between 2000 to 3000, the flow of liquid isunstable and may change from one type to the other.

85

4.4.6 Difference between streamline flow and turbulent flow

Str eamline flow Turbulent flow

1. The velocity of the fluid is 1. The velocity of the fluid issteady. unsteady.

2. The velocity of every 2. The velocity of every particleparticle crossing a particular crossing a particular point ispoint is the same. different.

3. The velocity of the fluid is 3. The velocity of the fluid isless than the critical velocity. greater than the critical velocity.

4. There is no mixing of various 4. There is always mixing oflayers with one another. various layers with one another.

Problem

1. What would be the maximum velocity of water in a pipe of diameter10 cm so that the flow is stream lined ? Coefficient of viscosityis 1 x 10-3 Nsm-2 .

For a streamline flow of maximum velocity, the Reynold number, K = 2000.

coefficient of viscosity η = 1 x 10 –3 Ns m–2

density of water ρ = 10 3 kg m-3

diameter of pipe D = 10 x 10-2 m

K = ρvD

η

v =Kη

=2000 x 1 x 10-3

ρD 103 x 10 x 10-2

maximum velocity v = 2 x 10-2 ms–1

86

4.5 Bernoulli’s theorem

Bernoulli’s theorem relates the velocity of a fluid at a point and thepressure of the fluid at that point. It is just the application of work-energytheorem. According to work energy theorem, the work done by a force

acting on a system is equal to the change in kinetic energy of the system.

Consider the streamlined flow of a liquid through a pipe as shown infig 4.10. As the liquid flows through the pipe, depending upon the position ofthe liquid, there are three types of energy possessed by the liquid during its

flow.

Kinetic energy

Let m be the mass of liquid that flows through the pipe with a velocity v.Kinetic energy of the liquid = mv2

Kinetic energy per unit mass of the liquid = v2

Potential energy

If h is the height from the ground, then the potential energy is given bymgh.

Potential energy per unit mass = gh

Pressure energy

If p is the pressure exerted on the liquid of cross sectional area a, then the

force acting on the liquid surface is given by

F = pa ( ... pressure = force / area)

Fig. 4.10 Bernoulli’s principle

12

2

h2

v2

p2

h1

v1

p1

87

Under the influence of this force, the liquid is driven through a smalldisplacement x, The work done is given by

w = Fx = p.a.xw = pV ( ... volume V = a x )

this work done is stored as the pressure energy.pressure energy = pV = p m/ρ

( ... density ρ = mass/ volume )pressure energy per unit mass = p/ρ

The three types of energy possessed by the liquid at two different points in thepipe (fig. 4.10) are as follows:

at A: potential energy per unit mass = gh1

kinetic energy per unit mass= v12

pressure energy per unit mass= p1/ρ

total energy at A = p1/ρ + gh

1 + v

12

at B: potential energy per unit mass= gh2

kinetic energy per unit mass= v22

pressure energy per unit mass= p2/ρ

total energy at B= p2/ρ + gh

2 + v

22

Bernoulli’ s theorem states that the sum of the energies possessed by aflowing liquid at any point is constant provided the flow of liquid is steady.

Fig. 4.11 Pressure energy

a

12

12

12

12

P

x

88

total energy at A = total energy at B

p1/ρ + gh

1 + v

12 = p

2/ρ + gh

2 + v

22

(ie) p/ρ + gh + v2 = constant

This is known as Bernoulli’s equation.

From the above equation, it is understood that when a fluid is in motion,the pressure within the fluid varies with the velocity of the fluid if the flow isstreamlined. The pressure within a fast moving fluid is lower than that in asimilar fluid moving slowly. This is known as Bernoulli’s principle.

4.5.1 Applications of the Bernoulli’s principle

1. Lift of an aeroplane The shape ofan aircraft wing is designed in such a waythat the velocity of the air above the wingis greater than that below it. A region oflow pressure is therefore created abovethe wing and so the air craft experiencesan upward force known as lift.

2. Atomiser or sprayer: It is used tospray liquid. When the rubber balloon ispressed, the air in the horizontal tubepasses with a large velocity near the nozzle.According to Bernoulli’s principle, thepressure in the tube will be reduced. Butthe pressure in the container is equal tothe atmospheric pressure. This pressuredifference makes the liquid rise in thevertical tube. The liquid is blown awaythrough the nozzle with a high velocity inthe form of fine spray.

Fig. 4.12 Aerofoil

1212

12

Fig. 4.13 Atomiser

low velocity,

high pressure

high velocity

low pressureair flow

Aerofoil

89

3) Bunsen burner

In a Bunsen burner, the gas comes out of the nozzle with high velocity.According to Bernoulli’s principle, the pressure in the stem of the burnerdecreases. So, air from the atmosphere rushes into the burner. The mixture of airand gas moves up the burner and burns.

Activity

1. Pour equal volumes of water and honey in two similar funnels. It will be

observed that water flows out of the funnel very quickly. On the other hand,

honey is slow in flowing down. This difference in the behaviour of two liquids

indicates that the viscosity of the two liquids have different values.

2. In order to visualize practically streamline and turbulent flow, take a

horizontal narrow glass tube. Place a small piece of soluble colouring matter

like crystals of potassium permanganate at one end of the tube. Connect this

end to a water tap.Open the tap so that water flows out of the tube slowly. A

coloured line is observed along the axis of the tube. The flow of liquid through

the tube is streamlined. Now open the tap fully. You will observe that the

colour spreads out through the entire liquid. The flow of liquid through the

tube now becomes turbulent.

air

Gas

Fig : 4.14. Bunsen burner

90

Let us muse upon

????? There are two types of solids namely crystalline and amorphous.

????? The external force which changes the relative position of the molecules ofthe body is called deforming force.

????? The property of the materials by virtue of which a body regains itsoriginal shape when the deforming force is removed is called elasticity.

????? The property by virtue of which the body does not regain its originalshape after the removal of deforming force is called plasticity.

????? The stress is defined as the restoring force developed per unit area ofthe body. Its unit is N m-2

????? The strain produced in a body is defined as the ratio of change indimension to its original dimension.

????? The maximum value of stress within which a body regains its originalshape is called elastic limit.

????? Hooke’s law states that the stress acting on a body is proportional tothe strain produced, within the elastic limit.

????? The force of attraction between the molecules of the same substanceis called cohesive force and the force of attraction between themolecules of different substances is called adhesive force.

????? The maximum distance upto which a molecule can exert an appreciable forceof attraction on another molecule is called molecular range.

????? Surface tension is defined as the force per unit length acting perpendicular toan imaginary line drawn on the liquid surface, tending to pull the surface apartalong the line.

????? The phenomenon of rise or fall of a liquid in a capillary tube is known ascapillarity.

91

????? Coefficient of viscosity of a liquid is numerically equal to the viscous forceacting tangentially between two layers of liquid having unit area of contactand unit velocity gradient normal to the direction of flow of the liquid.

????? Critical velocity of a liquid is the velocity below which the motion of the liquidis streamlined and above which the motion becomes turbulent.

????? The pressure within a fast moving fluid is lower than that in a similar fluidmoving slowly.

????? Reynold number is a mere number, which determines whether the flow ofliquid through a pipe is streamlined or turbulent.

Self evaluation

4.1 Intermolecular force of attraction exists between the moleculesseparated by a distance of abouta) 10-3 m b) 10-5 m c) 10-9 m d) 10-6 m

4.2 The restoring force per unit area of a deformed body is known asa) deforming force b) stress c) strain d) elastic limit

4.3 Bernoulli’s theorem is a consequence of the law of conservation ofa) momentum b) energy c) massd) angular momentum

4.4 An object entering the earth’s atmosphere at a high velocity catches firedue toa) viscosity of air b) the heat content of the atmospherec) the atmospheric pressure d) the higher force of gravity

4.5 For a streamlined flow, the value of Reynold number isa) above 3000 b) above 2000 c) below 2000 d) below 3000

92

4.6 The velocity of every liquid particle crossing a particular point insidea pipe is the same. The Reynold number for this type of flow will beranging from

a) 2000 to 3000 b) 0 to 2000 c) 2500 to 3500 d) 2000 to 2500

4.7 The unit of coefficient of viscosity is

a) Ns m-2 b) N s-1 m c) N s m-1 d) N s-2 m

4.8 Why are cotton dresses preferred in summer?

4.9 Why do small pieces of camphor dance about on the surface of water?

4.10 State Bernoulli’s principle.

4.11 On what factors does the rate of flow of a liquid through a pipedepend?

4.12 Explain molecular theory of surface tension.

4.13 Define coefficient of viscosity.

4.14 Distinguish between streamline and turbulent flow.

4.15 Explain the phenomenon of capillarity with suitable examples.

4.16 Explain a few applications of Bernoulli’s principle.

4.17 Why are bubbles emitted when a piece of chalk is immersed in water?

4.18 How do the insects run on the surface of water?

4.19 The nib of the pen is split. Why?

4.20 Why is oil poured to calm sea waves?

93

4.21 Explain the applications of surface tension in day–to-day life.

4.22 Water is flowing with a velocity of 15 cms–1 in a tube of diameter

1.1 cm. Find the Reynold number. Coefficient of viscosity of water is

10-3 N s m-2.

Answers

4.1 (c) 4.2 (b) 4.3 (b) 4.4 (a) 4.5 (c)

4.6 (b) 4.7 (a) 4.22 ( K = 1650)

94

5. MODERN PHYSICS

The nature of matter has been a subject of great speculation since earlytimes. It was felt that phenomena like evaporation, diffusion, emission,absorption of radiation and all chemical reactions could be explained only ifthe constituents of matter are known.

5.1 Atomic structur e

In an attempt to explain certain experimental facts about matter, variousatomic theories were suggested.

5.1.1 Dalton’s atomic theory

The first scientific theory about the constituents of matter was given byJohn Dalton in 1803. According to this theory matter consists of very smallparticles called atoms which cannot be sub-divided, destroyed or created.The atoms of the same substance are identical and differ from the atoms ofother substances. He further said that atoms retain their individuality in alltypes of chemical combinations.

5.1.2 Prout’s atomic theory

In 1815, Prout proposed that atoms of all elements were made up ofhydrogen atoms. Since many of the elements were found to have atomic weightsthat were not exact multiples of the hydrogen atom, this theory was discarded.

5.1.3 Electrical nature of matter

When Michael Faraday conducted experiments on passingelectricity through liquids, he established the electric nature of matter,i.e. matter is composed of positive and negative charges. Faraday hadno idea about the structure of the atom and was unaware of the origin ofthese positive and negative charges in matter.

95

5.1.4 Thomson’s model of an atom

J.J. Thomson in 1910 conducted an experiment by passing electricitythrough gases at reduced pressure. This experiment showed the existence ofnegatively charged particles in an atom. He named these particles as‘electrons’. It was during the same period the discovery of naturalradioactivity proved that an atom consists of equal quantities of positive andnegative charges, thus concluding that an atom is electrically neutral. Since thearrangement of the electrons in an atom remained unknown, as J.J. Thomsonattempted to give a structure to an atom.

According to Thomson’s atom model, an atom is a sphere of radius, about10–10m. The positive charges are uniformly distributed in the entire sphere andelectrons are embedded in it.

Since the distribution of electrons within the sphere of positive chargesresemble the plums in a pudding, J.J. Thomson’s atom model is referred to as thePlum-Pudding model.

5.1.5 Alpha - particle scattering

Based on the suggestion of Rutherford, Geiger and Marsden performed anexperiment on the scattering of α -particle from a thin gold foil. A schematicdiagram for the scattering of α - particles is shown in the figure given below:

Fig. 5.1 Thomson’s atom model

electron

positivelycharged sphere

96

A source of α - particle is placed in the cavity of a lead block. A narrow beamof α - particles is obtained by placing a slit in the path of the α - particles.

It is observed that most of the α - particles pass through the foilwithout any deviation showing that most of the space in the atom is empty.Some α - particles were found to scatter through a small angle and a fewα - particles were deflected through a large angle. A few turned back towardsthe source itself.

Based on the scattering of α - particle, the following inferences were made

i. The entire positive charge of the atom is concentrated in asmall portion at the centre of the atom called the nucleus.

ii. The space around the nucleus is almost empty.

5.1.6 Rutherford’s model of an atom

According to this model proposed by Rutherford in 1911 the entire positivecharge and the whole of its mass is concentrated in a tiny central core callednucleus.

The nucleus is surrounded by a suitable number of electrons such that the totalnegative charge is equal to the positive charge in the nucleus. These electronsrevolve in various circular orbits around the nucleus.

Fig. 5.2 Rutherford’s α - particles scattering experiment

slitlead blockgoldfoil

Sourceof α - particle

97

Rutherford’s assumed that the necessary centripetal force for the electrons torevolve around the nucleus is provided by the electrostatic force of attractionbetween the electrons and the nucleus.

5.1.7 Bohr’s model of an atom

Neil Bohr, a Danish physicist in 1913, presented an improved atom model.The entire concept of Bohr’s atom model is based on the following postulates.

Postulates of Bohr’s atom model

1. Electrons in an atom revolve around the nucleus under the action ofthe electrostatic force of attraction in a circular orbit.

2. An electron can revolve only in a few discrete and permitted orbits.While moving along these orbits an electron does not radiate energy. Thesenon-radiating orbits are called stationary orbits.

3. When electrons revolve in the permissible non-radiating orbit, theypossess a definite angular momentum (L = momentum x radius of the orbit)whose value is an integral multiple of h/2π, where h is Planck’s constant.

4. An atom will radiate energy only if the electron jumps from an orbit ofhigher energy (E

2) to another orbit of lower energy (E

1). The difference in

energy (E2 – E

1) determines the frequency of radiation.

i.e E2 – E

1 = h ν , where ν is called frequency of radiation.

Fig. 5.3 Rutherford’s atom model

nucleus

electron

98

Bohr could apply these postulates only for simple atoms like hydrogenand gave satisfactory explanation for the emission of spectral lines from ahydrogen atom. However, this atom model could not explain the fine structureof spectral lines.

5.2 Model of the nucleus

There were various models of the nucleus proposed by scientists fromtime to time. A brief description of a few such nuclear models are summarisedbelow.

5.2.1 Alpha particle model

In this model, the nucleus is supposed to have sub-groups in the form ofα- particles. Each sub-group has two protons and two neutrons.

5.2.2 Liquid drop model

According to this model, the nucleus is analogous in certain respects to thecharged liquid drop.The following analogy holds good between liquid dropand nucleus.

(i) A liquid drop is spherical due to surface tension and the nucleus is alsoassumed to be spherical in shape.

(ii) The splitting up of a liquid drop is similar to that of nuclear fission.

(iii) In the case of liquids, the intermolecular forces hold a liquid droptogether, and in the nucleus nuclear forces hold the nucleons (proton andneutron) together.

5.2.3 Nuclear structure

The atomic nucleus was discovered in 1911 by Rutherford and his associatesas a result of their experiment on scattering of α - particles by thin gold foils.Nucleus is a small positively charged sphere present at the centre of the atom.The radius of the nucleus is of the order of 10-14 m and is nearly 10,000 times

99

smaller than the radius of the atom. A nucleus accounts for almost the whole massof the atom. A study of various nuclear phenomena such as radioactivity revealsthat the nucleus is not a composite body but contains two types of particles, namelyneutrons and protons.

ProtonIt is an elementary particle. It has a positive charge equal in magnitude to the

charge of an electron i.e. 1.602 x 10-19 C. Its mass is 1.67 x 10-27 kg and is1836.1 times the mass of an electron.

NeutronIt is also an elementary particle. It is electrically neutral and has no charge. Its

mass is 1.675 x 10-27 kg and is 1838.6 times the mass of an electron. It is slightlyheavier than a proton.

Mass numberThe total number of nucleons (protons and neutrons) in the nucleus of the atom

is called the mass number of the atom and is denoted by A.

Atomic numberThe total number of protons in the nucleus of an atom is called the atomic

number of the atom and is denoted by Z.

If N represents the number of neutrons in the nucleus, then

Z + N = AN = A – Z

A nucleus is represented symbolically as Z X A, where X is the chemical symbol

of an element, Z is the atomic number and A is the mass number.

5.2.4 Properties of nucleus

(i) All nuclei are positively charged. The magnitude of the positive charge is anintegral multiple of the charge of an electron (ie) q = Ze where Z is the atomicnumber and e, the charge of an electron.

100

(ii) More than 99.9% of the mass of an atom is concentrated in the nucleus.The density of an atomic nucleus is very high and it shows that the nuclear matteris in a highly compressed state.

(iii) The distribution of the positive charge is uniform and the nucleus is spheri-cal in shape with a few exceptions.

(iv) The nuclear volume is found to be directly proportional to the total numberof nucleons i.e. mass number ‘A’.

i.e. volume α A

π r3 α A

r3 α A

r α A1/3

r = r0 A1/3

This is the empirical formula for nuclear radius, where r0 is the constant of

proportionality equal to 1.3x10-15m

5.3.1 Nuclear reaction

When the nucleus of an element is bombarded by an elementary particle or

a lighter nucleus, it undergoes a change in composition, producing one or

more nuclei of different elements. In this process, new lighter nuclei are formed

along with the emission of elementary particles and some amount of energy.

A nuclear reaction is symbolically represented by

A + x B + y + energy (Q)

From the above equation, it is understood that a nucleus A is bombarded

with an elementary particle ‘x’. As the result of this bombardment, a new

nucleus B is born along with the emission of another elementary particle

‘y’ and some energy Q.

43

101

Example

17Cl35 +

0 n1

15 P 32 +

2 He 4 + 0.935 MeV

17Cl35 is bombarded with a neutron to give

15 P 32 and an alpha particle with the

release of energy 0.935 MeV (1eV = 1.602 x 10–19

J and 1MeV = 106 eV).

Hence, a nuclear reaction is defined as the process of transformation ofan atomic nucleus into another nucleus by its interaction with an

elementary particle or a lighter nucleus.

5.3.2 Difference between nuclear reaction and chemical reaction.

Nuclear reaction Chemical reaction

1. A new element or isotope No new element is produced of an element is produced during a chemical reaction. during a nuclear reaction.

2. Nucleons take part in a nuclear Electrons take part in a chemical reaction. reaction.

3. It is not affected by any external It is affected by external agentsagents such as temperature, pressure,like temperature, pressure, etc. etc.

4. A nuclear reaction is always Both reversible and irreversible irreversible . reactions are possible in a

chemical reaction.

5. Enormous amount of energy is Energy released during chemical released during nuclear reaction. reaction is less.

5.3.3 Nuclear fission

In 1939, German scientists Otto Hahn and Strassman discovered that when a

92 U 235 nucleus is bombarded by thermal neutrons, it splits up into two smaller

nuclei of barium and krypton along with the release of three neutrons.

92 U 235+0 n 1 92 U 236 56 Ba 141 + 36 Kr 92+30 n1+Q

102

The figure 5.4 shows that a 92

U 235 nucleus undergoes distortion to attainthe shape of a dumb-bell and then splits into Ba 141 and Kr 92 with the releaseof three neutrons. The total mass of the initial particle is always greater thanthe total mass of the product obtained during nuclear fission. The differencein mass is converted into energy. The energy released per fission of

92 U 235

nucleus is found to be 200 MeV.

The phenomenon of breaking up of a heavy nucleus into two or morelighter nuclei of comparable masses with the release of large amountof energy is called nuclear fission.

Know it yourself

1. It is experimentally observed that barium and krypton were not the onlyproducts of nuclear fission of uranium. Depending on the kinetic energy ofthe neutrons, isotopes of different elements in the atomic number rangingfrom 34 to 58 are obtained as fission products.

2. When fast moving neutrons pass through matter they are slowed downby sharing their kinetic energy with the atoms of the matter due tocollision. This process of sharing of energy continues till the kineticenergy of the neutron is equal to the average kinetic energy of the atomsof the matter. The resulting neutron thus obtained is known as thermalneutron whose kinetic energy is less than 1 eV. Thermal neutron isefficient in causing a nuclear fission reaction.

Fig. 5.4 Nuclear fission

U235 U

236

n

n

n

Kr92

Ba141

n

103

Problem

Calculate the number of fissions that occur per second to produce a powerof 1 MW. Given that 200 MeV energy is released per fission of

92 U 235.

energy released per fission= 200 MeV = 200 x 10 6 eV

1 eV = 1.6 x 10 – 19 J

energy released per fission= 200 x 106 x 1.6 x 10-19 J

= 3.2 x 10 –11 J

energy required per second= 1 MW = 106 W

= 106 J s–1

number of fissions per second=energy required

energy released per fission

= 106

3.2 x 10-11

number of fissions= 3.125 x 1016

per second

Know it yourself

Naturally occurring uranium consists of two isotopes. About 0.7% is

U235 and 99.3% is U238. Experiments indicate that 92

U 235 is a

fissionable material.

5.3.4 Nuclear chain reactionWhen

92 U 235 nucleus splits up, it generally releases three neutrons which are

likely to produce nuclear fission of three more 92

U 235 nuclei. On an average aneutron per fission is either absorbed by non-fissionable material or escapeswithout hitting any other

92 U 235 nuclei . The remaining two neutrons hit

two other nuclei and produce some more neutrons and so on. The numberof fissions taking place at each successive stage goes on increasing ingeometrical progression.

104

Nuclear chain reaction is a process in which if initially fission isinduced in a single fissile nucleus, then all other fissile nuclei willautomatically undergo fission by increasing the number of neutrons ingeometrical progression.

Reproduction factor (k) It is an important parameter in nuclear fissionreaction that determines whether the fission reaction is controlled or uncontrolled.It is defined as the ratio of the number of neutrons present in a particulargeneration to the number of neutrons present in the precedinggeneration.

From the figure 5.5, it is seen that the number of neutrons in the first generationis 1 and that in the beginning of the second generation is 2. Now the reproduction

factor k = 2.

For an uncontrolled chain reactionk > 1

For a controlled chain reaction k = 1

For the reaction to stop k < 1

Fig. 5.5 Nuclear chain reaction

Neutron

105

5.3.5 Critical size

For a sustained nuclear chain reaction to take place, the size of the fuel is veryimportant. If the size of the fuel is too small, then the liberated neutrons escapefrom the surface without producing further fission and the growth of thechain reaction ends. Therefore, there must be a minimum size of the nuclearfuel below which the chain reaction stops. This size is called critical size. Thecorresponding mass is called critical mass.

5.4 Uncontrolled chain reaction

Once a chain reaction is set up in a certain mass (above the critical mass),it will accelerate at a very fast rate and an enormous amount of energy is releasedin a very short time. Such an uncontrolled chain reaction forms the basicprinciple of an atom bomb.

Atom bomb

Two hemispheres of the same fissionable material U235or Pu239 each ofmass lesser than critical mass are kept apart by a distance as shown in thefigure 5.6. When the bomb is to be exploded a third piece of U235 in the form ofa cylinder is propelled, so that it will fuse together with the other two pieces.Since the total mass now is greater than the critical mass of the fissionablesubstance, the uncontrolled chain reaction takes place.

During the time of explosion, the temperature rises to millions of degree celsiusand the pressure rises to millions of atmosphere. The explosion is accompaniedby a violent and intense blast of visible light, ultra violet, and x-rays causing ablinding flash. In addition, a large amount of radioactive radiations and neutronsare also released.

Fig. 5.6 Atom bomb

sub-criticalmasses

Remote controlledchemical explosive

U235

106

5.5 Controlled chain reactionThe chain reaction is so vigorous and of such a short duration that the

tremendous amount of energy released can cause a heavy destruction.The chain reaction can, however, be controlled by absorbing excess ofneutrons released in the process of nuclear fission, so that on an average, aneutron from each fission reaction is left to produce further fission. Underthese conditions, the number of fissions produced per second remainsconstant and the energy released does not go out of control. This energy can,therefore, be used for peaceful purposes. Such a chain reaction is called acontrolled chain reaction.

5.5.1 Nuclear reactor

A nuclear reactor is a device in which the nuclear fission reaction is carriedon as a perfectly controlled chain reaction, in a self-sustained manner. Themain parts of a nuclear reactor are explained below.

Nuclear fuel

It is the fissionable material required for the fission process to take place.Commonly used fuels in the reactors are U235, Pu239 and U 233.

Neutron sourceA suitable neutron source is conveniently arranged to start the fission

reaction in the reactor.

ModeratorIt is found that the slow moving neutrons (thermal neutrons) are efficient in

producing fission reaction. Moderators are used to slow down the fastmoving neutrons. Heavy water or graphite is used as a moderator.

Control rodsIn order to ensure that the rate of fission that takes place in the reactor is

constant, the excess neutrons are absorbed with the help of control rods.Cadmium or boron strip is used as control rods.

107

Coolant

The enormous amount of heat energy released during fission is absorbed bycoolants. Heavy water is used as a coolant. The heat absorbed by the coolantis transferred to the water in the heat exchanger. Due to this heat exchange,superheated steam is produced which rotates the turbines to generate electricpower.

Neutron reflector

In order to prevent the escape of neutrons, a reflector is fixed around the fueland the moderator.

Radiation shieldingIn order to prevent the leakage of nuclear radiations, the reactor is usually

surrounded by a thick lead lining which in turn is surrounded by a concrete wall ofthickness of about 2 to 2.5 m.

Uses

A nuclear reactor is generally used for the following purposes.

1. To generate electric power.2. To produce radio isotopes.3. To produce a neutron beam of high intensity which is used in nuclear

research and in the treatment of cancer.4. To produce radioactive plutonium from U238, which is required for explosive

producing explosion.

Fig. 5.7 Nuclear reactor

Control rods

To turbineShielding

Fuel

Coolant To turbine

108

5.5.2 Reactors in IndiaAfter the establishment of the Atomic Energy Commission, India has built nuclear

reactors at BARC, Mumbai. Apsara, Cirus, , Dhruva and Purnima are a fewresearch purpose reactors. Kamini , at IGCAR (Indira GandhiCentre for Atomic Research), Kalpakkam is another research reactor.

The operating nuclear power reactors in India are at Kalpakkam (Tamilnadu),Kakrapara (Gujarat), Kaiga (Karnataka), Tarapur (Maharashtra) andKota (Rajasthan).

Fast breeder reactorIn this type of reactor, fission is caused by fast neutrons and thus no moderator

is required. Here both fissionable (92

U235) and fertile (92

U238) materials can beused. The fast neutrons are absorbed by

92U238 and gets converted into Pu239,

which is a fissionable material. These types of reactors, which produce theirown fuel during their operation, are called fast breeder reactors. We haveone breeder reactor in Kalpakkam.

5.5.3 Nuclear fusion

When two or more lighter nuclei fuse together to form a heavynucleus, an enormous amount of energy is released. This process is callednuclear fusion.

In a fusion reaction, the mass of heavier nucleus formed is less than the totalmass of lighter nuclei. The difference in mass is fully converted into energy ,according to Einstein’s mass - energy relation, E = mc2, where m is the massdifference and c is the velocity of light. A nuclear fusion reaction takes placeonly at a very high temperature of about 107 K. The hydrogen bomb is basedon the principle of the nuclear fusion reaction.

Example 1H2 +

1H3

2He4 +

0n1 + energy

When deuterium fuses with tritium, helium is formed. As a result, a neutron is

released accompanied by an energy of 17.6 MeV.

109

Let us muse upon

????? Prout proposed that atoms of all elements were made up of hydrogen

atom.

????? J.J. Thomson first showed the existence of negatively charged particles

(electrons) in an atom.

????? The radius of an atom is found to be about 10-10 m.

????? J.J. Thomson’s atom model is referred as the plum-pudding model.

????? The scattering of α - particle on a thin gold foil shows that positively

charged matter is concentrated at the centre in a very small region called

nucleus.

????? Rutherford discovered the nucleus of an atom.

????? The radius of the nucleus is of the order of 10-14 m and is nearly 10,000

times smaller than the radius of the atom.

????? Mass number of an atom is defined as the sum of the number of protons

and neutrons.

????? The total number of protons in the nucleus of an atom is called the atomic

number.

????? The magnitude of the electric charge of the nucleus is an integral multiple

of the charge of an electron. (q = Z e)

????? The nuclear volume is found to be directly proportional to the total

number of nucleons. (mass number)

????? Nuclear reaction is defined as the process of transformation of an atomic

nucleus into another nucleus by its interaction with an elementary

particle or a lighter nucleus.

110

????? The phenomenon of breaking up of a heavy nucleus into two or more

lighter nuclei of comparable masses is called nuclear fission.

????? Nuclear chain reaction is a process in which, if initially fission is induced

in a single fissile nucleus, then all other fissile nuclei will automatically

undergo fission by increasing the number of neutrons in geometrical

progression.

????? The ratio of the number of neutrons present in a particular generation in achain reaction to the number of neutrons present in the precedinggeneration is called reproduction factor.

????? The reactors which produce their own fuel during their operation arecalled breeder reactors.

????? The process in which two or more light nuclei fuse together to form aheavy nucleus is called nuclear fusion.

Self evaluation

5.1 According to Prout’s model, the atoms of all elements are made up of

(a) protons (b) protons and neutrons

(c) protons, neutrons and electrons(d) hydrogen atom

5.2 The radius of an atom is about

(a) 10 –14 m (b) 10 – 12 m (c) 10 – 10 m (d) 10 – 15 m

5.3 Large angle scattering of α-particles shows that

(a) an atom is a positively charged sphere

(b) a positive charge is concentrated only at the centre of an atom.

(c) atom has no electrons

(d) negative charge is concentrated only at the centre of an atom.

5.4 The nuclear radius is

(a) 10 –14 m (b) 10 – 10 m (c) 10 – 9 m (d) 10 – 7 m

111

5.5 Nuclear radius is _________ times smaller than the radius of the atom.(a) 100 (b) 1000 (c) 10,000 (d) 2000

5.6 Electrons are embedded in a positively charged sphere. This wassuggested by

(a) Thomson (b) Rutherford (c) Bohr (d) Dalton

5.7 The mass of a neutron is

(a) 9.11 x 10 –31 kg (b) 1.66 x 10 – 27 kg

(c) 1.675 x 10– 27 kg (d) 1.656 x 10 – 29 kg

5.8 The formula for nuclear radius is

(a) r = A ro1/3 (b) r = r

0 A1/2 (c) r = r

0 A1/3 (d) r = r

0 A-1/3

5.9 A chain reaction is possible when the mass of the fuel is greater than(a) proton mass (b) neutron mass

(c) electron mass (d) critical mass

5.10 A controlled chain reaction takes place in

(a) atom bomb (b) nuclear reactor

(c) hydrogen bomb (d) all of these

5.11 The coolant used in a nuclear reactor is

(a) graphite (b) uranium (c) heavy water (d) water

5.12 What is Prout’s atom model?

5.13 What is alpha-particle model of a nucleus?

5.14 Define mass number and atomic number.

5.15 Define nuclear fission.

5.16 What is critical mass?

5.17 What is a nuclear reaction? Give an example.

5.18 Describe Rutherford’s α – particles scattering experiment.

5.19 Explain Bohr’s model of an atom.

112

5.20 Give the properties of a nucleus.

5.21 Write a short note on an atom bomb.

5.22 Write a note on chain reaction.

5.23 Explain the liquid drop model of a nucleus.

5.24 Define reproduction factor.

5.25 Explain the essential parts of a nuclear reactor.

5.26 Distinguish between nuclear fission and fusion.

5.27 Find the radius of a nucleus of mass number 64.

5.28 If the energy released per fission is 200 MeV, calculate the number of

fissions that takes place per second to produce a power of 10 kW.

Answers

5.1 (d) 5.2 (c) 5.3 (b) 5.4 (a) 5.5 (c)

5.6 (a) 5.7 (c) 5.8 (c) 5.9 (d) 5.10(b)

5.11 (c) 5.27. (5.2 x 10–15

m) 5.28. (3.125 x 1014

)

113

6. X - RAYS AND RADIOACTIVITY

X - rays were accidentally discovered by W.C Roentgen in 1895. He namedthis radiation as X - rays because its nature and properties could not be known atthat time. Almost one year after the discovery of X - rays, radioactivity wasdiscovered by Henry Becquerel in 1896.

The discovery of natural radioactivity marked the beginning of one of themost fruitful developments in modern physics. A detailed study of it hasbeen of utmost importance in atomic physics forming the basis of many highlysuggestive discoveries about an atom, such as the nuclear atom model, isotopicconstitution, artificial disintegration, induced radioactivity, etc. In this chapter, weshall discuss the principle, the discovery, properties and uses of X - rays and alsothe basic concepts on natural and artificial radioactivity.

6.1 X- RAYS

When W.C. Roentgen was experimenting with

the discharge of electricity through gases, he

covered the discharge tube with a thin black card

board to facilitate better observation of

fluorescence. He accidentally noticed a brilliant glow

on a screen coated with barium platino cyanide

lying in a corner of his laboratory. He

concluded that the invisible radiations coming out

from the discharge tube should have caused the

fluorescence on the screen. Roentgen called these

invisible penetrating rays which affected the screen

as X - rays.

The discovery of these rays in 1895 has opened a new field of study inmedicine, surgery and industry and many forms of X-ray tubes have beendesigned to suit their needs.

W.C. Roentgen

114

6.1.1 Production of X - rays

Principle X - rays are produced when fast moving electrons aresuddenly stopped by a metal target of high atomic mass.

The modern X-ray tube was designed by Dr.William Coolidge in 1912.

It consists of a glass tube which is highly evacuated up to 10-7 m of mercury.

A tungsten filament coated with barium oxide acts as a cathode. When

heated by current from the low tension battery, it emits electrons. It is surrounded

by a molybdenum cylinder to focus the electrons. Tungsten which has a high

melting point acts as an anode and is inclined at an angle of 450. It is embedded

in a copper block. Copper fins are provided at the end of the copper block

and water is circulated through pipes to cool the anode.

A very high potential of about 100,000V from the secondary of an induction

coil is applied between the anode and the cathode This high potential

accelerates the electrons. When these electrons strike the tungsten target,

X - rays are produced.

Fig 6.1 – Production of X - rays

tungsten filament

molybdenumcylinder

L T. battery X - rays

Copper fins

target tungsten

115

6.1.2. Properties of X - rays

1. X-rays are electromagnetic radiations of very short wavelength of the order of 10–10m.

2. As they are not charged particles, they are not deflected by electric and magnetic fields.

3. They produce fluorescence in materials coated with barium platino- cyanide and zinc sulphide.

4. They travel in straight lines with the velocity of light.

5. They affect photographic plates.

6. They ionise the gases through which they pass.

7. They are photons of high energy.

8. They are diffracted by crystals.

9. They cause chemical and biological changes.

10. They have high penetrating power, but are stopped by lead and bones.

6.1.3. Applications of X - rays

MEDICAL 1. X-rays can pnetrate through flesh and not throughbones. Hence they are used to detect (a) fractures anddislocation of bones (b) diseased organs and unwantedgrowth of bones and stones in the body (c) the presenceof foreign bodies like bullets, pins and coins in thehuman body.

2. X - rays of very high penetrating power are used todestroy malignant tumours and cure skin diseases.

INDUSTRY X - rays are used to detect

1. crack in metal structures - the body of aeroplanesand automobiles.

2. the quality of welding in moulds and metal castings.

3. the presence of pearls in oysters.

116

DETECTION They are used to detect the smuggling of precious metals,explosives and contraband goods.

RESEARCH X - rays are used to study the structure of crystals, which iscalled X - ray crystallography.

Know it yourself

The harmful effects of X-rays were known almost as soon as they were

discovered. An X-ray photograph is taken when we cannot physically

see or check an organ. When X-rays pass through the human body a

part of it is absorbed and some part goes right through.

In the human body there are three parts which are affected by

X-rays. First, is the genitalia, which, when affected may have a

negative effect on the progeny. The is the skin. X - rays can cause

rashes, hair loss and can lead to cancer. The third is the red blood cells

which, when affected can cause anaemia. If the white blood cells are

affected they can attack the immune system and lead to various

diseases.

The International Commission on Radiological protection has

recommended that the effective dose for a member of the public shall

not exceed 1mSv per year averaged over a period of 5 years with not

more than 5 mSv in any given year during the five year period.

We express radiation dosage in a unit called the “sievert (Sv)”. Smaller

quantities are expressed in millisievert. (mSv)

Average dose for dental X - ray is 0.1 mSv and chest X - ray

is 0.08 mSv

117

6.2. Radioactivity

In 1896, a French physicistProf. Henri Becquerel, discovered that aphotographic plate, lying near a uraniumcompound was affected. This prompted himto carry out the experiment with other saltsof uranium and he concluded thaturanium and its salts emit invisibleradiations which can pass through paper,wood, glass etc. and affect photographicplates.

Natural radioactivity is the spontaneous emission of alpha, beta, andgamma rays by the nuclei of heavy elements whose atomic number isgreater than 82. These elements are called ‘radioactive elements’.

6.2.1 Rutherford’ s experiment - Natural radioactivity

Henry Becquerel

Fig 6.2 Rutherford’s experiment

lead blockradioactive material

magnetic field

photographic plate

to vacuum pump

118

The nature of the radioactive radiationswere studied experimentally by Dr. ErnestRutherford.

A small quantity of a radioactiveelement is placed in the cavity of a leadblock. A photographic plate is placed closeto it. A strong magnetic field isset up perpendicular to the plane of thepaper directed inwards and the wholearrangement is placed in an evacuatedchamber.

On examining the photographic plate, three traces are observed indicating thatthere are three types of radiations. They are alpha particles, beta rays and gammarays.

According to Fleming’s left hand rule, the particles which get deflectedtowards the left are positively charged and are called alpha particles.The rays which get deflected towards the right are negatively charged and arecalled beta rays. The rays which are undeviated are gamma rays.

6.2.2. Properties of radioactive radiations

ALPHA RAYS BETA RAYS GAMMA RAYS

1.Alpha particles are Beta rays are electrons Gamma rays arehelium nuclei electromagnetic

radiations

2.They are positively They are negatively They do not havecharged particles charged particles any charge.

Dr. Ernest Rutherford

119

3.They are deflected onlyThey are deflected by They are not

by strong electric and electric and magnetic deflected by electricmagnetic fields. fields. or magnetic field.

4.Their velocity is much Being light particles their Their velocity is

less than the velocity velocity is nearly equal tothe same as theof light and is between the velocity of light. velocity of light.

1.4 x 107 m s–1 and2.15 x 107 m s–1.

5.Their ionising power Their ionising power is They have a very

is very high. less. low ionising power.

6.They affect They affect photographicThey affectphotographic plates. plates. photographic plates.

7.When they are incidentWhen they are incident When they are incidenton materials coated withon materials coated on materials coated

zinc sulphide they causewith zinc sulphide,they with zinc sulphide,fluorescence. cause fluorescence they cause fluorescence

8.Their penetrating powerThey have a higher They have a very high

is very low. They can bepenetrating power. penetrating power.stopped by cardboard orThey can be stopped by They can be stopped

aluminium of 0.01 cm aluminium of 0.5cm or only by lead of 30cmthickness. lead of 1mm thickness. thickness.

9.When an alpha particleWhen a beta particle is When a gamma ray

is emitted by the nucleus,emitted, the mass is emitted there is nothe mass number number remains the change in the mass

decreases by 4, and thesame, but the atomic number or atomicatomic number decreasesnumber increases number.

by 2. by one.

120

6.3. Forces of nature

Physicists have classified the basic forces of nature into four types such asgravitational, electromagnetic, strong and weak nuclear forces.

a) Gravitational force

(i). It is that force which keeps the earth orbiting the sun and holds the galaxies together.

(ii). It is the weakest of all the known forces, yet it dominates the universe.(iii). It is a long range force(iv). It is an attractive force.

b) Electromagnetic force (i). It is the force between any two electric charges or two current carrying

conductors. (ii). It is stronger than gravitational force.(iii). It is either attractive or repulsive.

c) Str ong nuclear force (i). It is the force that holds nucleons together. It is the force between

two nucleons i.e. the force between two protons (p-p), two neutrons (n-n) or between a neutron and a proton (n-p). (ii). It is the strongest of all forces.(iii). It acts over short distances.(iv). It is attractive and charge independent.

d) Weak nuclear force (i). It is the force that arises in nuclear processes such as beta decay.(ii). It is a very weak force(iii). The range of the force is shorter than the size of the protons or neutrons.

121

6.4. Ar tificial radioactivity

In 1934, Irene Curie and herhusband Joliot found that even lighter nucleielements which are stable could bemade radioactive by artificial means whenthey were bombarded by high energyparticles such as alpha particles.

When aluminium is bombarded byalpha particles an isotope of radioactive phosphorous is produced, with therelease of a neutron.

13A l27

+

2He4 –––––>

15P30 +

0n1

When boron is bombarded by alpha particles radioactive nitrogen isproduced with the release of a neutron.

5B10 +

2He4 –––––>

7N13 +

0n1

Artificial or induced radioactivity is the transmutation of anonradioactive element into a radioactive element by artificial means.These artificially produced radioactive elements are called radio-isotopes.

This discovery led to the production of more than 800 radio-isotopes which donot occur in nature.

6.4.1. Dif ferences between natural and artifical radioactivity .

Natural radioactivity Artificial radioactivity

1. Natural radioactive elements Radioactive isotopes are not found in are found in nature. nature but are produced by artificial means.

2. Only heavy elements whose Artificial radioactivity is exhibited by both atomic number is greater than light and heavy elements. 82 exhibit natural radioactivity.

3. Positrons are not emitted. Positrons are emitted. Positrons arepositively charged particles whose mass is equal to the mass of an electron.

Irene Curie

122

6.4.2. Uses of Radio-isotopes

Radio-isotopes have provided a powerful tool for the solution of numerous

problems in biology, physiology, chemistry and other sciences in addition to

industry. Some of the problems particularly those connected with life processes

could not have been solved without the use of isotopes. A few applications of

radio-isotopes are discussed below.

1. In medicine

(a) In diagnosis - Radio iodine 53

I131 is used in determining the condition

of the thyroid gland. Radio sodium 11

Na24 and Radio potassium 19

K42 are used to

detect disorders in blood circulation. Radio chromium 24

Cr51 is used to locate

the exact position of haemorrhage in the blood circulatory system.

(b) In therapy - Radio cobalt 27

Co60 is used in the treatment of cancer.

Radio iron 26

Fe59 is used to treat anaemia. Radio Gallium 31

Ga67 is used to treat

soft tissue tumours. Radio Strontium 38

Sr90 is used to treat skin cancer.

2. In industry - Special paints mixed with isotopes of Promethium 61

Pm147

are used for illuminating watches, aircraft instrument dials, safety signs,

self - luminous paints and post markers.

3. In agriculture - Radio isotopes are used in the production and

preservation of agricultural products. The shelf - life of edible items can

be increased by irradiation. Sprouting in onion, garlic, potatoes and ginger

can be reduced by exposing the containers to gamma rays from 27

Co60 for a

certain duration. It also slows down the ripening of fruits like bananas, mangoes

and papayas. Radio phosphorus 15

P32 is used in the soil as a fertilizer

and the absorption of phosphorus by the plant from the soil is studied.

123

Radio-carbon dating

It is a technique of estimating the ages of mummies, wooden implements, etc,

using radioactive carbon (6C14) present in them. Radio carbon is produced in the

atmosphere by the bombardment of 7N14 nuclei by neutrons present in

cosmic rays. The half life period of 6C14 is 5600 years. Half life period of a

radioactive element is defined as the time taken for half the number ofits atoms to disintegrate. Living organisms take in

6C14 with their food they

eat and the air they breathe in. Intake of 6C14 stops at their death. Thereafter

the radiocarbon present in them decreases due to its decay. By measuring the

percentage of 6C14 present in relics, their ages are calculated.

Let us muse upon

F X - rays are produced when fast moving electrons are suddenly stopped bya metal target of high mass number.

F X - rays can pass through human body but not through bones.

F X - rays are used to study the structure of crystals.

F The spontaneous emission of alpha, beta and gamma rays by the nucleus ofheavy elements whose atomic number is greater than 82 is calledradioactivity.

F The ionising power of alpha particles is very high.

F The penetrating power of gamma rays is very high.

F When a radioactive nucleus emits an alpha particle the atomic number ofthe nucleus decreases by 2 and its mass number decreases by 4.

F When a radioactive nucleus emits a beta ray, the atomic number increases by one and there is no change in the mass number.

F When a radioactive nucleus emits gamma rays, there is no change in theatomic number and mass number of the nucleus.

124

F There are four types of basic forces namely gravitational force, electromagnetic force, strong and weak nuclear forces.

F The gravitational force is the weakest of all the known forces.

F Artificial or induced radioactivity is the transmutation of a non-radioactiveelement into a radioactive element.

F The age of relics is estimated by a technique called radiocarbon dating.

Self evaluation

6.1. Natural radio activity was first observed bya) Henri Becquerel b) Rutherford c) Joliot d) Irene Curie.

6.2. The transmutation of a non-radioactive element into a radioactiveelement is called

a) natural radioactivity b) artificial radioactivityc) chain reaction d) nuclear fusion

6.3. X - rays were discovered bya) Rutherford b) Coolidgec) W.C. Roentgen d) Madam Curie

6.4. The production of X - rays requiresa) high voltage and low pressure b) high voltage and high pressurec) low voltage high pressure d) low voltage and low pressure

6.5. The slowest among the following area) alpha particles b) beta rays c) gamma rays d) X - rays

6.6. The radiations of the maximum ionising power area) X - rays b) beta rays c) gamma rays d) alpha particles

6.7. The weakest among the basic forces of nature isa) electro magnetic force b) gravitational forcec) weak nuclear force d) strong nuclear force

6.8. Natural radioactivity occurs in elements of atomic number greater thana)28 b) 82 c) 52 d) 40

125

6.9. Among the following the force that is charge independent isa)electrostatic force b) electromagnetic forcec) nuclear force d) all the above forces

6.10. The radio isotope used to treat anaemia in the human body isa) Ga67 b) Fe59 c) Sr90 d)Co60

6.11. What are the basic forces of nature?

6.12. Define natural radioactivity.

6.13. What is artificial radioactivity?

6.14. What is the principle involved in producing X-rays?

6.15. Compare the penetrating and ionising properties of alpha, beta andgamma rays.

6.16. Distinguish between natural and artificial radioactivity.

6.17. Explain induced radioactivity with examples.

6.18. Mention the appliciations of radioisotopes in the field of medicines.

6.19. How are radio isotopes used in agriculture?

6.20. How are radio isotopes used for industrial purposes?

6.21. Write the uses of X-rays.

6.22. Mention the properties of X-rays.

6.23. With the help of a neat diagram explain the production of X-rays.

6.24. Describe Rutherford’s experiment on radioactivity.

Answers

6.1. (a) 6.2. (b) 6.3. (c) 6.4. (a) 6.5. (a)

6.6. (d) 6.7. (b) 6.8.(b) 6.9.(c) 6.10.(b)

126

7. UNIVERSE

The science which deals with the study of heavenly bodies with respect to

their motion, position and composition is known as astronomy. Astrophysics

deals with the physical properties and interaction of celestial bodies, the application

of the principles of physics to celestial bodies and their phenomena. Traditionally

astronomy has been more observational in scope while Astrophysics is more

analytical. Thus, astronomers look through telescopes while astrophysicists

analyse the collected data.

7.1 Planetary motion

Long before the rise of civilization, primitive man who was mainly nomadic in

culture was using the sun and stars for finding the direction during his journeys.

The earliest evidence of planetary observations appeared around 1600 B.C.

Ptolemy, a Greek astronomer studied the motion of the planets in detail and

proposed a theory known as Geocentric theory. According to this theory, the

earth is at the centre of the universe and all other heavenly bodies move around

it in various orbits. Aryabhatta, the Indian astronomer was the first to describe

the rotation of the earth on its own axis. He also ascertained the fact that the

earth rotates west- east by citing various illustrations.

The 16th century A.D marked the transition from mediaeval to modern

astronomy by two great advances: the suggestion of a heliocentric system by

Copernicus and the systematic recording of accurate observations of Tycho Brahe.

According to Heliocentric theory the sun is at the centre and all the planets

revolve around it in circular orbits. Based on the observations made by

Tycho Brahe, a German astronomer, Johannes Kepler proposed the three

famous empirical laws of planetary motion.

127

7.2 Kepler’s Laws of Planetary Motion

First Law (Law of orbits)

Each planet moves around the sun in an elliptical orbit with the sun atone of its foci.

P is a planet revolving round the sunas shown in fig. 7.1. The closestposition A of the planet from the sun iscalled perigee and the farthest position Bof the planet from the sun is called apogee.

Second Law (Law of areas)

The line joining the sun and the planet sweeps out equal areas in equalintervals of time.

The orbit of the planet around the sun isas shown in the fig. 7.2. In a given intervalof time, the planet covers longer distanceA to B near the perigee and short distanceC to D near the apogee. Hence, the speedof the planet is maximum at the perigeeposition and minimum at the apogeeposition.

Third law (Law of periods)

The square of the period of revolution of a planet (T) around the sunis directly proportional to the cube of the mean distance between theplanet and the sun (r).

(i.e.) T2 α r

3

T2

r3

Planetary data applied to Kepler’s third law are tabulated below. It can be seen

that the ratio is constant for all the planets.

Fig. 7.1

= constant

T2

r3

A B

Sun

SunA

B C

D

Fig. 7.2

P

128

TABLE 7.1

Name of the Time period Mean distance (x10-25)Planet T (years) from the sun r years2

(x 109m) km3

Mercury 0.241 57.90 2.991

Venus 0.615 108.21 2.985

Earth 1.000 149.60 2.987

Mars 1.881 227.94 2.988

Jupiter 11.862 778.30 2.985

Saturn 29.458 1427.00 2.986

Uranus 84.015 2869.00 2.990

Neptune 164.788 4498.00 2.984

Pluto 248.400 5900.00 3.004

7.3 Newton’s universal law of gravitation

Every body in the universe attracts every other body with a force whichis directly proportional to the product of the masses of the two bodies andinversely proportional to the square of the distance between them.

If m1 and m

2 are the masses of two bodies separated by a distance r, the force

of attraction F between them is given by

F α m1 m

2

F α

T2

r3

1

r 2

Fig. 7.3

r

129

F α F =

where, G is a constant known as the universal gravitational constant.

The value of G = 6.67 x 10 –11 N m 2 kg–2

If m1 = m

2 = 1 kg and r = 1m, then F = G.

Thus, the gravitational constant is numerically equal to the force ofattraction between two bodies each of mass 1 kg separated by a distanceof 1 metre.

7.4 The solar system

Solar system is the part of the universe in which the sun occupies the central

position of the system holding together all the heavenly bodies such as planets,

moons, minor planets or asteroids, comets and meteors. The sun holds all other

heavenly bodies under the influence of the gravitational force. You might have

studied about the nine planets in your earlier classes viz. Mercury, Venus, Earth,

Mars, Jupiter, Saturn, Uranus, Neptune and Pluto.

Asteroids

These are huge lumps of rocks orbiting round the sun mainly between Marsand Jupiter. They are also called minor planets and are considered to be theremains of a planet which broke up long ago. Ceres is the largest knownasteroid.

Comets

Comets are insignificant tiny rock like objects with frozen gases movingaround the sun in elliptical orbits having the sun at one of its foci.

During its motion, the comets pass quite close to the sun and then recede faraway into the inter-planetary space. When comets travel quite far away from thesun, gaseous materials accumulate and freeze on the icy bodies. When theyapproach the sun, the solar heat vaporizes the condensed matter and are

m1 m

2

r2

G m1 m

2

r2

130

pushed away from the sun by the solar wind. Thus comets develop a tail pointingaway from the sun. Some comets are seen at fixed regular intervals of time.Halley’s Comet is a periodic comet which makes its appearance once in seventy

six years. Its last appearance was in 1986 and it would appear again in 2062.

Meteors and meteorites

Meteors also called as shooting stars are rocky and metallicparticles or fragments revolving round the sun. These may be the fragmentsof the asteroids or comets. When these pieces enter the earth’s atmosphere theyare burnt up by the heat generated due to friction in the earth’s atmosphere. As aresult we see a streak of light. The streak vanishes within a few seconds becausethe rocky pieces melt and vaporize completely in a short time.

Some bigger pieces may survive the heat produced by friction andthey manage to reach the earth and they are called as meteorites.

7.5 Galaxy

A cluster of stars, gas and dust particles held togetherby gravitational for ce is called a galaxy. Millions of galaxies havebeen photographed by the modern telescopes and they constitute theuniverse. Galaxies consist of millions of stars. Our Galaxy - the Milky wayis spiral in shape. The nearest galaxy to us is Andromeda galaxy which is at adistance of 2 x 106 light year. (The distance travelled by light in one year iscalled a light year. One light year = 9.467 x 1012 km.)

Expansion of universe

An American astronomer Edwin Hubble observed that all galaxies are rushingaway from each other. He established the relationship between the speed (V)and the distance(R) of the galaxies. The velocity with which galaxies arereceding away is proportional to the distance between them.

i.e. V α R

V = HR where, H is Hubble’s constant

This is known as Hubble’s law

131

7.6 Origin of the universe

The Big bang theory

According to this theory all matter in the universe was concentrated to a smallcore which was dense and hot in the beginning. Some cosmic explosion occurredaround 15 billion years ago. As a result the matter was thrown out in alldirections in the form of galaxies and they were receding away from each other.The presence of cosmic radiation and observation of Edwin Hubble that thegalaxies are moving away from each other supports this theory. As the galaxiesmove further and further all the matter in the galaxy will get used up and the

universe will be empty at one point of time.

There are also two other theories namely- Pulsating theory and Steady

State theory that explain the origin of the universe.

7.7 Ar tificial satellites

A body moving in an orbit around a planet is called a satellite.The moon is the natural satellite of the earth. Ar tificial satellites areman-made objects, which revolve around the earth. These satellites are

used for communication purpose, weather monitoring etc.

Satellites used for communication purpose have a time period of 24 hours

(i.e) the same as that of our earth. These satellites are called geo - stationary

satellites. Everything that happens in the world comes to our door - step only

with the help of communication satellites.

Remote sensing

Collection of information about an object without physical contactwith the object is known as remote sensing. The remote sensing satellites

can be used in agriculture, forestry, drought assessment, estimation of crop fields,

detection of fishing zones, mapping and surveying.

132

Indian space mission

The Indian Government established the Department Of Space (DOS) in

1972 to promote development and application of space science and technology

for socio - economic benefits. Indian Space Research Organisation (ISRO), is

the primary agency under Department of Space, for executing space programmes.

During the seventies, India undertook demonstration of space applications for

communication, broadcasting and remote sensing; designing and building

experimental satellites - Aryabhatta, Bhaskara, APPLE and Rohini, and

experimental Satellite Launch Vehicles - SLV - 3 and ASLV. India has

established space systems that form an important element of the national

infrastructure.

Today, it is clear that the resolution has paid rich dividends, since, India

has emerged as a front-ranking nation, not only in the development of space

technology, but also in taking its benefits to the people. The Indian National

Satellite (INSAT) and Indian Remote Sensing (IRS) satellite systems form the

important components for the development of the infrastructure for

telecommunication, television and broadcasting, meteorology, disaster warning

and resource management for our country.

Early experiments

Aryabhatta The first Indian Satellite was launched into a near earth orbit

on April 19, 1975 by an Intercosmos rocket of the erstwhile Russia, for conductingscientific experiments in space and to gain experience in the design, fabricationand operation of a complete system.

Bhaskara - I & II They carried two TV cameras, one in visible and theother in near - infrared band, and a 3 - frequency passive microwave radiometer.

APPLE (Ariane Passenger Payload Experiment) It was used to conductseveral communication experiments.

133

Rohini series While the first Rohini satellite was used to measure theperformance parameters of SLV 3, the second and third carried landmarksensor payloads viz. Stretched Rohini Series Satellites, SROSS C andSROSS C2.

Satellite telecommunication experiment project (STEP) providedexperience in the operation of geostationary satellite system for domestictelecommunication in designing and building ground infrastructure.

Space science The research under the Indian space programme encompassesa wide spectrum of activities including a study of cosmic rays, astronomicalinvestigation using space and ground - based systems, study of meteorites, lunarsamples and physical observations of the sun. Another major branch of activityrelates to the study of earth’s atmospheric system through rockets, balloons andorbiting space systems.

Know it yourself

Some of the space centres in India are listed below:Vikram Sarabhai Space Centre (VSSC), Thiruvanathapuram,

pioneers in rocket research and in planning and execution of launch vehicledevelopment projects of ISRO.

SRO Satellite Centre (ISAC), Bangalore is responsible for the design,fabrication, testing and management of satellite systems for scientific,technological and application missions.

Space Applications Centre (SAC), Ahmedabad. The major fieldsof activity cover satellite communications, remote sensing andmeteorology.

SHAR Centre, Sriharikota. This centre also undertakes large scaleproduction of solid rocket propellant and ground testing of solid fuelledrocket stages of launch vehicle.

National Remote Sensing Agency (NRSA), Hyderabad, underDOS, has facilities for surveying, identifying, classifying and monitoringearth resources using aerial and satellite data.

134

Know it yourself TABLE 7.2

NAME OF THE DATE OF LAUNCH LAUNCHEDSATELLITE LAUNCH VEHICLE AT

ARYABHATA 19.04.1975 INTER COSMOS RUSSIA

BHASKARA – 1 07.06.1979 INTER COSMOS RUSSIA

RS – 1 18.07.1980 SLV – 3 INDIA

RS – D1 31.05.1981 SLV – 3 INDIA

APPLE 19.06.1981 ARIANE FRANCE

BHASKARA – 2 20.11.1981 INTER COSMOS RUSSIA

INSAT – 1A 10.04.1982 US DELTA USA

RS – D2 17.04.1983 SLV – 3 INDIA

INSAT – 1B 30.08.1983 SPACE SHUTTLE USA

SROSS – 1 24.03.1987 ASLV INDIA

IRS – 1A 17.03.1988 VOSTOK RUSSIA

SROSS – 2 13.07.1988 ASLV INDIA

INSAT – 1C 22.07.1988 ARIANE FRANCE

INSAT – 1D 12.06.1990 US DELTA USA

IRS – IB 29.08.1991 VOSTOK RUSSIA

SROSS – C 20.05.1992 ASLV INDIA

INSAT – 2A 10.07.1992 ARIANE FRANCE

INSAT – 2B 23.07.1993 ARIANE FRANCE

IRS – 1E 20.09.1993 PSLV INDIA

SROSS – C2 04.05.1994 ASLV INDIA

IRS – P2 15.10.1994 PSLV INDIA

INSAT – 2C 07.12.1995 ARIANE FRANCE

IRS – 1C 28.11.1995 MOLNIA RUSSIA

IRS – P3 21.03.1996 PSLV INDIA

INSAT – 2D 04.06.1997 ARIANE FRANCE

IRS – 1D 29.09.1997 PSLV – C1 INDIA

135

INSAT – 2E 03.04.1999 ARIANE FRANCE

IRS – P4 26.05.1999 PSLV – C2 INDIA

INSAT – 3B 22.03.2000 ARIANE FRANCE

GSAT – 1 18.04.2001 GSLV – D1 INDIA

INSAT – 3C 24.01.2002 ARIANE FRANCE

KALPANA – 1 12.09.2002 PSLV – C4 INDIA

INSAT – 3A 10.04.2003 ARIANE FRANCE

GSAT – 2 08.05.2003 GSLV – D2 INDIA

IRS – P6 17.10.2003 PSLV – C5 INDIA

EDUSAT 20.09.2004 GSLV – F01 INDIA

CARTOSAT – 1 &

HAMSAT 05.05.2005 PSLV – C6 INDIA

Let us muse upon

F The branch of science which deals with the study of heavenly bodies is

known as Astronomy.

According to the geocentric theory, the earth is at the centre of the universe.

According to the heliocentric theory, the sun is at the centre and the earthrotates on its axis around the sun.

Kepler’s laws

(i) Each planet moves around the sun in an elliptical orbit with the sun atone of its foci.

(ii) The line joining the sun and the planet sweeps out equal areas in equalintervals of time.

(iii) The square of the period of the revolution of a planet around the sun is

directly proportional to the cube of the mean distance between the planet andthe sun.

The value of universal gravitational constant G = 6.67 x 10 – 11 Nm2 kg – 2

F

F

F

F

136

Asteroids are found between the orbit of Mars and Jupiter.

A cluster of stars is called a galaxy.

The nearest galaxy is Andromeda galaxy.

The shape of our Milky Way galaxy is spiral.

The distance travelled by light in one year is called a light year.

Man made objects that revolve around the earth are called artificial satellites.

Collection of information about an object without physical contact with itis known as remote sensing.

The first satellite launched by India was Aryabhatta.

The time period of geostationary satellites is 24 hours which is same as that of

our earth.

Comets develop a tail as they approach the sun.

Self evaluation

7.1 The geo-centric theory was proposed by(a) Ptolemy (b) Copernicus(c) Johannes Kepler (d) Newton

7.2 According to the law of periods(a) V α r (b) T α r (c) T3 α r2 (d) T2 α r3

7.3 The value of universal gravitational constant G is(a) 6.76 x 10 – 11Nm2 kg – 2 (b) 6.67 x 10 – 11 Nm – 2kg – 2

(c) 6.67 x 10 – 11 Nm2 kg – 2 (d) 6.67 x 10 11 Nm 2 kg – 2

7.4 The unit of G is ____________ (a) ms – 2 (b) Nm 2 (c) Nm2 kg – 2 (d) Nm – 2kg – 2

7.5 The objects which are found between orbits of Mars and Jupiter are(a) meteors (b) meteorites (c) comets (d) asteroids

F

F

F

F

F

F

F

F

F

F

137

7.6 The shape of our galaxy is(a) elliptical (b) spiral (c) irregular (d) none of the above

7.7 What is helio-centric theory?

7.8 What is geo-centric theory?

7.9 State Kepler’s laws of planetary motion.

7.10 State the Universal law of gravitation.

7.11 Two students seated on a bench do not feel the gravitational force ofattraction. Why?

7.12 What are asteroids?

7.13 Why is the moon not used for communication purpose?

7.14 Write a note on : a. Comet b. Meteor c. Meteorite

7.15 State Hubble’s law.

7.16 What is a galaxy?

7.17 What are the uses of artificial satellites?

7.18 What is remote sensing?

7.19 What are called geostationary satellites?

7.20 Explain Big-bang theory.

7.21 Define light year.

7.22 Calculate the gravitational force of attraction between two bodies of masses

400 kg and 50 kg separated by a distance of 2 m.

Answers

7.1 (a) 7.2 (d) 7.3 (c) 7.4 (c) 7.5 (d)

7.6 (b) 7.22 3.335 x 10 – 7 N

163

LOGARITHMS

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

Mean Differences

1 2 3 4 5 6 7 8 90 1 2 3 4 5 6 7 8 9

0000

0414

0792

1139

1461

1761

2041

2304

2553

2788

3010

3222

3424

3617

3802

3979

4150

4314

4472

4624

4771

4914

5051

5185

5315

5441

5563

5682

5798

5911

6021

6128

6232

6335

6435

6532

6628

6721

6812

6902

6990

7076

7160

7243

7324

4 8 12 17 21 25 29 33 37

4 8 11 15 19 23 26 30 34

3 7 10 14 17 21 24 28 31

3 6 10 13 16 19 23 26 29

3 6 9 12 15 18 21 24 27

3 6 8 11 14 17 20 22 25

3 5 8 11 13 16 18 21 24

2 5 7 10 12 15 17 20 22

2 5 7 9 12 14 16 19 21

2 4 7 9 11 13 16 18 20

2 4 6 8 11 13 15 17 19

2 4 6 8 10 12 14 16 18

2 4 6 8 10 12 14 15 17

2 4 6 7 9 11 13 15 17

2 4 5 7 9 11 12 14 16

2 3 5 7 9 10 12 14 15

2 3 5 7 8 10 11 13 15

2 3 5 6 8 9 11 13 14

2 3 5 6 8 9 11 12 14

1 3 4 6 7 9 10 12 13

1 3 4 6 7 9 10 11 13

1 3 4 6 7 8 10 11 12

1 3 4 5 7 8 9 11 12

1 3 4 5 6 8 9 10 12

1 3 4 5 6 8 9 10 11

1 2 4 5 6 7 9 10 11

1 2 4 5 6 7 8 10 11

1 2 3 5 6 7 8 9 10

1 2 3 5 6 7 8 9 10

1 2 3 4 5 7 8 9 10

1 2 3 4 5 6 8 9 10

1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 7 8

1 2 3 4 5 5 6 7 8

1 2 3 4 4 5 6 7 8

1 2 3 4 4 5 6 7 8

1 2 3 3 4 5 6 7 8

1 2 3 3 4 5 6 7 8

1 2 2 3 4 5 6 7 8

1 2 2 3 4 5 6 6 7

1 2 2 3 4 5 6 6 7

0043

0453

0828

1173

1492

1790

2068

2330

2577

2810

3032

3243

3444

3636

3820

3997

4166

4330

4487

4639

4786

4928

5065

5198

5328

5453

5575

5694

5809

5922

6031

6138

6243

6345

6444

6542

6637

6730

6821

6911

6998

7084

7168

7251

7332

0086

0492

0864

1206

1523

1818

2095

2355

2601

2833

3054

3263

3464

3655

3838

4014

4183

4346

4502

4654

4800

4942

5079

5211

5340

5465

5587

5705

5821

5933

6042

6419

6253

6355

6454

6551

6646

6739

6830

6920

7007

7093

7177

7259

7340

0128

0531

0899

1239

1553

1847

2122

2380

2625

2856

3075

3284

3483

3674

3856

4031

4200

4362

4518

4669

4814

4955

5092

5224

5353

5478

5599

5717

5832

5944

6053

6160

6263

6365

6464

6561

6656

6749

6839

6928

7016

7101

7185

7267

7348

0170

0569

0934

1271

1584

1875

2148

2405

2648

2878

3096

3304

3502

3692

3874

4048

4216

4378

4533

4683

4829

4969

5105

5237

5366

5490

5611

5729

5843

5955

6064

6170

6274

6375

6474

6571

6665

6758

6848

6937

7024

7110

7193

7275

7356

0212

0607

0969

1303

1614

1903

2175

2430

2672

2900

3118

3324

3522

3711

3892

4065

4232

4393

4548

4698

4843

4983

5119

5250

5378

5502

5623

5740

5855

5966

6075

6180

6284

6385

6484

6580

6675

6767

6857

6946

7033

7118

7202

7284

7364

0253

0645

1004

1335

1644

1931

2201

2455

2695

2923

3139

3345

3541

3729

3909

4082

4249

4409

4564

4713

4857

4997

5132

5263

5391

5514

5635

5752

5866

5977

6085

6191

6294

6395

6493

6590

6684

6776

6866

6955

7042

7126

7210

7292

7372

0294

0682

1038

1367

1673

1959

2227

2480

2718

2945

3160

3365

3560

3747

3927

4099

4265

4425

4579

4728

4871

5011

5145

5276

5403

5527

5647

5763

5877

5988

6096

6201

6304

6405

6503

6599

6693

6785

6875

6964

7050

7135

7218

7300

7380

0334

0719

1072

1399

1703

1987

2253

2504

2742

2967

3181

3385

3579

3766

3945

4116

4281

4440

4594

4742

4886

5024

5159

5289

5416

5539

5658

5775

5888

5999

6107

6212

6314

6415

6513

6609

6702

6794

6884

6972

7059

7143

7226

7308

7388

0374

0755

1106

1430

1732

2014

2279

2529

2765

2989

3201

3404

3598

3784

3962

4133

4298

4456

4609

4757

4900

5038

5172

5302

5428

5551

5670

5786

5899

6010

6117

6222

6325

6425

6522

6618

6712

6803

6893

6981

7067

7152

7235

7316

7396

164

LOGARITHMS

55

56

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

Mean Differences

1 2 3 4 5 6 7 8 90 1 2 3 4 5 6 7 8 9

7404

7482

7559

7634

7709

7782

7853

7924

7993

8062

8129

8195

8261

8325

8388

8451

8513

8573

8633

8692

8751

8808

8865

8921

8976

9031

9085

9138

9191

9243

9294

9345

9395

9445

9494

9542

9590

9638

9685

9731

9777

9823

9868

9912

9956

1 2 2 3 4 5 5 6 7

1 2 2 3 4 5 5 6 7

1 2 2 3 4 5 5 6 7

1 1 2 3 4 4 5 6 7

1 1 2 3 4 4 5 6 7

1 1 2 3 4 4 5 6 6

1 1 2 3 4 4 5 6 6

1 1 2 3 3 4 5 6 6

1 1 2 3 3 4 5 5 6

1 1 2 3 3 4 5 5 6

1 1 2 3 3 4 5 5 6

1 1 2 3 3 4 5 5 6

1 1 2 3 3 4 5 5 6

1 1 2 3 3 4 4 5 6

1 1 2 2 3 4 4 5 6

1 1 2 2 3 4 4 5 6

1 1 2 2 3 4 4 5 5

1 1 2 2 3 4 4 5 5

1 1 2 2 3 4 4 5 5

1 1 2 2 3 3 4 4 5

1 1 2 2 3 3 4 4 5

1 1 2 2 3 3 4 4 5

1 1 2 2 3 3 4 4 5

1 1 2 2 3 3 4 4 5

1 1 2 2 3 3 4 4 5

1 1 2 2 3 3 4 4 5

1 1 2 2 3 3 4 4 5

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 4 4

0 1 1 2 2 3 3 3 4

7412

7490

7566

7642

7716

7789

7860

7931

8000

8069

8136

8202

8267

8331

8395

8457

8519

8579

8639

8698

8756

8814

8871

8927

8982

9036

9090

9143

9196

9248

9299

9350

9400

9450

9499

9547

9595

9643

9689

9736

9782

9827

9872

9917

9961

7419

7497

7574

7649

7723

7796

7868

7938

8007

8075

8142

8209

8274

8338

8401

8463

8525

8585

8645

8704

8762

8820

8876

8932

8987

9042

9096

9149

9201

9253

9304

9355

9405

9455

9504

9552

9600

9647

9694

9741

9786

9832

9877

9921

9965

7427

7505

7582

7657

7731

7803

7875

7945

8014

8082

8149

8215

8280

8344

8407

8470

8531

8591

8651

8710

8768

8825

8882

8938

8993

9047

9101

9154

9026

9258

9309

9360

9410

9460

9509

9557

9605

9653

9699

9745

9791

9836

9881

9926

9969

7435

7513

7589

7664

7738

7810

7882

7952

8021

8089

8156

8222

8287

8351

8414

8476

8537

8597

8657

8716

8774

8831

8887

8943

8998

9053

9106

9159

9212

9263

9315

9365

9415

9465

9513

9562

9609

9657

9703

9750

9795

9841

9886

9930

9974

7443

7520

7597

7672

7745

7818

7889

7959

8028

8096

8162

8228

8293

8357

8420

8482

8543

8603

8663

8722

8779

8837

8893

8949

9004

9058

9112

9165

9217

9269

9320

9370

9420

9469

9518

9566

9614

9661

9708

9754

9800

9845

9890

9934

9978

7451

7528

7604

7679

7752

7825

7896

7966

8035

8102

8169

8235

8299

8363

8426

8488

8549

8609

8669

8727

8785

8842

8899

8954

9009

9063

9117

9170

9222

9274

9325

9375

9425

9474

9523

9571

9619

9666

9713

9759

9805

9850

9894

9939

9983

7459

7536

7612

7686

7760

7832

7903

7973

8041

8109

8176

8241

8306

8370

8432

8494

8555

8615

8675

8733

8791

8848

8904

8960

9015

9069

9122

9175

9227

9279

9330

9380

9430

9479

9528

9576

9624

9671

9717

9763

9809

9854

9899

9943

9987

7466

7543

7619

7694

7767

7839

7910

7980

8048

8116

8182

8248

8312

8376

8439

8500

8561

8621

8681

8739

8797

8854

8910

8965

9020

9074

9128

9180

9232

3284

9335

9385

9435

9484

9533

9581

9628

9675

9722

9768

9814

9859

9903

9948

9991

7474

7551

7627

7701

7774

7846

7917

7987

8055

8122

8189

8254

8319

8382

8445

8506

8567

8627

8686

8745

8802

8859

8915

8971

9025

9079

9133

9186

9238

9289

9340

9390

9440

9489

9538

9586

9633

9680

9727

9773

9818

9863

9908

9952

9996

165

ANTILOGARITHMS

.00

.01

.02

.03

.04

.05

.06

.07

.08

.09

.10

.11

.12

.13

.14

.15

.16

.17

.18

.19

.20

.21

.22

.23

.24

.25

.26

.27

.28

.29

.30

.31

.32

.33

.34

.35

.36

.37

.38

.39

.40

.41

.42

.43

.44

.45

.46

.47

.48

.49

Mean Differences

1 2 3 4 5 6 7 8 90 1 2 3 4 5 6 7 8 9

1000

1023104710721096

1122

1148117512021230

1259

1288131813491380

1418

1445147915141549

1585

16221660169817381778

18201862190519501995

20422089213821882239

22912344239924552512

25702630269227542818

2884295130203090

0 0 1 1 1 1 2 2 2

0 0 1 1 1 1 2 2 20 0 1 1 1 1 2 2 20 0 1 1 1 1 2 2 20 1 1 1 1 2 2 2 2

0 1 1 1 1 2 2 2 2

0 1 1 1 1 2 2 2 20 1 1 1 1 2 2 2 20 1 1 1 1 2 2 2 30 1 1 1 1 2 2 2 3

0 1 1 1 1 2 2 2 3

0 1 1 1 2 2 2 2 30 1 1 1 2 2 2 2 30 1 1 1 2 2 2 3 30 1 1 1 2 2 2 3 3

0 1 1 1 2 2 2 3 3

0 1 1 1 2 2 2 3 30 1 1 1 2 2 2 3 30 1 1 1 2 2 2 3 30 1 1 1 2 2 3 3 3

0 1 1 1 2 2 3 3 3

0 1 1 2 2 2 3 3 30 1 1 2 2 2 3 3 30 1 1 2 2 2 3 3 40 1 1 2 2 2 3 3 40 1 1 2 2 2 3 3 4

0 1 1 2 2 2 3 3 40 1 1 2 2 2 3 3 40 1 1 2 2 2 3 4 40 1 1 2 2 2 3 4 40 1 1 2 2 2 3 4 4

0 1 1 2 2 3 3 4 40 1 1 2 2 3 3 4 40 1 1 2 2 3 3 4 41 1 2 2 3 3 4 4 51 1 2 2 3 3 4 4 5

1 1 2 2 3 3 4 4 51 1 2 2 3 3 4 4 51 1 2 2 3 3 4 4 51 1 2 2 3 3 4 5 51 1 2 2 3 3 4 5 5

1 1 2 2 3 4 4 5 51 1 3 3 3 4 4 5 61 1 2 3 3 4 4 5 61 1 2 3 3 4 4 5 61 1 2 3 3 4 5 5 6

1 1 2 3 3 4 5 5 61 1 2 3 3 4 5 5 61 1 2 3 4 4 5 6 61 1 2 3 4 4 5 6 6

1002

1026105010741099

1125

1151117812051233

1262

1291132113521384

1416

1449148315171552

1589

16261663170217421782

18241866191019542000

20462094214321932244

22962350240424602518

25762636269827612825

2891295830273097

1005

1028105210761102

1127

1153118012081236

1265

1294132413551387

1419

1452148615211556

1592

16291667170617461786

18281871191419592004

20512099214821982249

23012355241024662523

25822642270427672831

2897296530343105

1007

1030105410791104

1130

1156118312111239

1268

1297132713581390

1422

1455148915241560

1596

16331671171017501791

18321875191919632009

20562104215322032254

23072360241524722529

25882649271027732838

2904297230413112

1009

1033105710811107

1132

1159118612131242

1271

1300133013611393

1426

1459149315281563

1600

16371675171417541795

18371879192319682014

20612109215822082259

23122366242124772535

25942655271627802844

2911297930483119

1012

1035105910841109

1135

1161118912161245

1274

1303133413651396

1429

1462149615311567

1603

16411679171817581799

18411884192819722018

20652113216322132265

23172371242724832541

26002661272327862851

2917298530553126

1014

1038106210861112

1138

1164119112191247

1276

1306133713681400

1432

1466150015351570

1607

16441683172217621803

18451888193219772023

20752118216822182265

23232377243224892547

26062667272927932858

2924299230623133

1016

1040106410891114

1140

1167119412221250

1279

1309134013711403

1435

1469150315381574

1611

16481687172617661807

18491892193619822028

20752123217322232275

23282382243824952553

26122673273527992864

2931299930693141

1019

1042106710911117

1143

1169119712251253

1282

1312134313741406

1439

1472150715421578

1514

16521690173017701811

18541897194118962032

20802128217822282280

23332388244325002559

26182679274228052871

2938300630763148

1021

1045106910941119

1146

1172119912271256

1285

1315134613771409

1442

1476151015451581

1618

16561694173417741816

18581901194519912037

20842133218322342286

23392393244925062564

26242685274828122877

2944301330833155

166

ANTILOGARITHMS

.50

.51

.52

.53

.54

.55

.56

.57

.58

.59

.60

.61

.62

.63

.64

.65

.66

.67

.68

.69

.70

.71

.72

.73

.74

.75

.76

.77

.78

.79

.80

.81

.82

.83

.84

.85

.86

.87

.88

.89

.90

.91

.92

.93

.94

.95

.96

.97

.98

.99

Mean Differences

1 2 3 4 5 6 7 8 90 1 2 3 4 5 6 7 8 9

3162

3236331133883467

3548

3631371538023890

3981

4074416942664365

4467

4571467747864898

5012

51295248537054955623

57545888602661666310

64576607676169187079

72447413758677627943

81288318851187108913

9120933395509772

1 1 2 3 4 4 5 6 7

1 2 2 3 4 5 5 6 71 2 2 3 4 5 5 6 71 2 2 3 4 5 6 6 71 2 2 3 4 5 6 6 7

1 2 2 3 4 5 6 7 7

1 2 3 3 4 5 6 7 81 2 3 3 4 5 6 7 81 2 3 4 4 5 6 7 81 2 3 4 5 5 6 7 8

1 2 3 4 5 6 6 7 8

1 2 3 4 5 6 7 8 91 2 3 4 5 6 7 8 91 2 3 4 5 6 7 8 91 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 9 101 2 3 4 5 7 8 9 101 2 3 4 6 7 8 9 101 2 3 4 6 7 8 9 10

1 2 3 4 6 7 8 9 11

1 2 4 5 6 7 8 10 111 2 4 5 6 7 9 10 111 3 4 5 6 8 9 10 111 3 4 5 6 8 9 10 121 3 4 5 7 8 9 10 12

1 3 4 5 7 8 9 11 121 3 4 5 7 8 10 11 121 3 4 6 7 8 10 11 131 3 4 6 7 9 10 11 131 3 4 6 7 9 10 12 13

2 3 5 6 8 9 11 12 142 3 5 6 8 9 11 12 142 3 5 6 8 9 11 13 142 3 5 6 8 10 11 13 152 3 5 6 8 10 11 13 15

2 3 5 7 8 10 12 13 152 3 5 7 9 10 12 14 162 4 5 7 9 11 12 14 162 4 6 7 9 11 13 14 162 4 6 7 9 11 13 15 17

2 4 6 8 9 11 13 15 172 4 6 8 10 12 14 15 172 4 6 8 10 12 14 16 182 4 6 8 10 12 14 16 182 4 6 8 10 12 15 17 19

2 4 6 8 11 13 15 17 192 4 7 9 11 13 15 17 202 4 7 9 11 13 16 18 202 5 7 9 11 14 16 18 20

3170

3243331933963475

3556

3639372438113899

3990

4083417842764375

4477

4581468847974909

5023

51405260538355085636

57685902603961806324

64716622677669347096

72617430760377807962

81478337853187308933

9141935495729795

3177

3251332734043483

3565

3648373338193908

3999

4093418842854385

4487

4592469948084920

5035

51525272539555215649

57815916605361946339

64866637679269507112

72787447762177987980

81668356855187508954

9162937695949817

3184

3258333434123491

3573

3656374138283917

4009

4102419842954395

4498

4603471048194932

5047

51645284540855345662

57945929606762096353

65016653680869667129

72957464763878167998

81858375857087708974

9183939496169840

3192

3266334234203499

3581

3664375038373926

4018

4111420743054406

4508

4613472148314943

5058

51765297542055465675

58085943608162236368

65166668682369827145

73117482765678348017

82048395859587908995

9204941996389863

3199

3273335034283508

3589

3673375838463936

4027

4121421743154416

4519

4624473248424955

5070

51885309543355595689

58215957609562376383

65316683683969987161

73287499767478528035

82228414861088109016

9226944196619886

3206

3281335734363516

3597

3681376738553945

4036

4130422743254426

4529

4634474248534966

5082

52005321544555725702

58345970610962526397

65466699685570157178

73457516769178708054

82418433863088319036

9247946296839908

3214

3289336534433524

3606

3690377638643954

4046

4140423643354436

4539

4645475348644977

5093

52125333545855855715

58485984612462666412

65616714687170317194

73627534770978898072

82608453865088519057

9265948497059931

3221

3296337334513532

3614

3698378438733963

4055

4150424643454446

4550

4656476448754989

5105

52245346547055985728

58615998613862816427

65776730688770477211

73797551772779078091

82798472867088729078

9290950697279954

3228

3304338134593540

3622

3707379338823972

4064

4159425643554457

4560

4667477548875000

5117

52365358548356105741

58756012615262956442

65926745690270637228

73967568774579258110

82998492869088929099

9311952897509977

167

NATURAL SINES

00

1

2

3

4

5

6

7

8

9

100

11

12

13

14

15

16

17

18

19

200

21

22

23

24

25

26

27

28

29

300

31

32

33

34

35

36

37

38

39

400

41

42

43

44

1 ’ 2 ’ 3 ’ 4 ’ 5 ’

0.0000

0.0175

0.0349

0.0523

0.0698

0.0872

0.1045

0.1219

0.1392

0.1564

0.1736

0.1908

0.2079

0.2250

0.2419

0.2588

0.2756

0.2924

0.3090

0.3256

0.3420

0.3584

0.3746

0.3907

0.4067

0.4226

0.4384

0.4540

0.4695

0.4848

0.5000

0.5150

0.5299

0.5446

0.5592

0.5736

0.5878

0.6018

0.6157

0.6293

0.6428

0.6561

0.6691

0.6820

0.6947

3 6 9 12 15

3 6 9 12 15

3 6 9 12 15

3 6 9 12 15

3 6 9 12 15

3 6 9 12 14

3 6 9 12 14

3 6 9 12 14

3 6 9 12 14

3 6 9 12 14

3 6 9 11 14

3 6 9 11 14

3 6 9 11 14

3 6 8 11 14

3 6 8 11 14

3 6 8 11 14

3 6 8 11 14

3 6 8 11 14

3 6 8 11 14

3 5 8 11 14

3 5 8 11 14

3 5 8 11 14

3 5 8 11 14

3 5 8 11 14

3 5 8 11 13

3 5 8 11 13

3 5 8 11 13

3 5 8 11 13

3 5 8 11 13

3 5 8 11 13

3 5 8 11 13

2 5 7 10 12

2 5 7 10 12

2 5 7 10 12

2 5 7 10 12

2 5 7 9 12

2 5 7 9 12

2 5 7 9 12

2 5 7 9 11

2 5 7 9 11

2 5 7 9 11

2 4 7 9 11

2 4 6 9 11

2 4 6 8 11

2 4 6 8 10

0017

0192

0366

0541

0715

0889

1063

1236

1409

1582

1754

1925

2096

2267

2436

265

2773

2940

3107

3272

3473

3600

3762

3923

4083

4242

4399

4555

4710

4863

5015

5165

5314

5461

5606

5750

5892

6032

6170

6307

6441

6574

6704

6833

6959

0035

0209

0384

0558

0732

0906

1080

1253

1426

1599

1771

1942

2113

2284

2453

2622

2790

2957

3123

3289

3453

3616

3778

3939

4099

4258

4415

4571

4726

4879

5030

5180

5329

5476

5621

5764

5906

6046

6184

6320

6455

6587

6717

6845

6972

0052

0227

401

0576

0750

0924

1097

1271

1444

1616

1788

1959

2130

2300

2470

2639

2807

2974

3140

3305

3469

3633

3795

3955

4115

4274

4431

4586

4741

4897

5045

5195

5344

5490

5635

5779

5920

6060

6198

6334

6468

6600

6730

6858

6984

0070

0244

0419

0593

0767

0941

1115

1288

1461

1633

1805

1977

2147

2317

2487

2656

2823

2990

3156

3322

3486

3649

3811

3971

4131

4289

4446

4602

4756

4909

5060

5210

5358

5505

5650

5793

5934

6074

6211

6347

6481

6613

6743

6871

6997

0087

0262

0436

0610

0785

0958

1132

1305

1478

1650

1822

1994

2164

2334

2504

2672

2840

3007

3173

3338

3502

3665

3827

3987

4147

4305

4462

4617

4772

4924

5075

5225

5373

5519

5664

5807

5948

6088

6225

6361

6494

6626

6756

6884

7009

0105

0279

0454

0628

0802

0976

1149

1323

1495

1668

1840

2011

2181

2351

2521

2689

2857

3024

3190

3355

3518

3681

3843

4003

4163

4321

4478

4633

4787

4939

5090

5241

5388

5534

5678

5821

5962

6101

6239

6374

6508

6639

6769

6896

7022

0122

0297

0471

0645

0819

0993

1167

1340

1513

1685

1857

2028

2198

2368

2538

276

2874

3040

3206

3371

3535

3697

3859

4019

4179

4337

4493

4648

4802

4955

5105

5255

5402

5548

5693

5835

5976

6115

6252

6388

6521

6652

6782

6909

7034

0140

0314

0488

0663

0837

1011

1184

1357

1530

1702

1874

2045

2215

2385

2554

2723

2890

3057

3223

3387

3551

3714

3875

4035

4195

4352

4509

4664

4818

4970

5120

5270

5417

5563

5707

5850

5990

6129

6266

6401

6534

6665

6794

6921

7046

0157

0332

0506

0680

0854

1028

1201

1374

1547

1719

1891

2062

2233

2402

2571

2740

2907

3074

3239

3404

3567

3730

3891

4051

4210

4368

4524

4679

4833

4985

5135

5284

8432

5577

5271

5864

6004

6143

6280

6414

6547

6678

6807

6934

7059

0’

0.00

6’

0.10

12’

0.20

18’

0.30

24’

0.40

30’

0.50

36’

0.60

42’

0.70

48’

0.80

54’

0.90

Mean Differences

168

NATURAL SINES

450

46

47

48

49

500

51

52

53

54

55

56

57

58

59

600

61

62

63

64

65

66

67

68

69

700

71

72

73

74

75

76

77

78

79

800

81

82

83

84

85

86

87

88

89

Mean Differences

1 ’ 2 ’ 3 ’ 4 ’ 5 ’

0.7071

0.7193

0.7314

0.743

1.7547

0.7660

0.7771

0.7880

0.7986

0.8090

0.8192

0.8290

0.8387

0.8480

0.8572

0.8660

0.8746

0.8829

0.8910

0.8988

0.9063

0.9135

0.9205

0.9272

0.9336

0.9397

0.9455

0.9511

0.9563

0.9613

0.9659

0.9703

0.9744

0.9781

0.9816

0.9848

0.9877

0.9903

0.9925

0.9945

0.9962

0.9976

0.9986

0.9994

0.9998

2 4 6 8 10

2 4 6 8 10

2 4 6 8 10

2 4 6 8 10

2 4 6 8 9

2 4 6 7 9

2 4 5 7 9

2 4 5 7 9

2 3 5 7 9

2 3 5 7 8

2 3 5 7 8

2 3 5 6 8

2 3 5 6 8

2 3 5 6 8

1 3 4 6 7

1 3 4 6 7

1 3 4 6 7

1 3 4 5 7

1 3 4 5 6

1 3 4 5 6

1 2 4 5 6

1 2 3 5 6

1 2 3 4 6

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 4 5

1 2 3 3 4

1 2 2 3 4

1 2 2 3 4

1 1 2 3 4

1 1 2 3 3

1 1 2 3 3

1 1 2 2 3

1 1 2 2 3

0 1 1 2 2

0 1 1 2 2

0 1 1 2 2

0 1 1 1 2

0 1 1 1 2

0 0 1 1 1

0 0 1 1 1

0 0 0 1 1

0 0 0 0 0

0 0 0 0 0

7083

7206

7325

7443

7559

7672

7782

7891

7997

8100

8202

8300

8396

8490

8581

8669

8755

8838

8918

8996

9070

9143

9212

9278

9342

9403

9461

9516

9568

9617

9664

9707

9748

9785

9820

9851

9880

9905

9928

9947

9963

9977

9987

9995

9999

7096

7218

7337

7455

7570

7683

7793

9702

8007

8111

8211

8310

846

8499

8590

8678

8763

8846

8926

9003

9078

9150

8219

9285

9348

9409

9466

9521

9573

9622

9668

9711

9751

9789

9823

9854

9882

9907

9930

9949

9965

9978

9988

9995

9999

7108

7230

7249

7466

7581

7694

7804

7912

8018

8121

8221

8320

8415

8508

8599

8686

8771

8854

8934

9011

9085

9157

9225

9291

9354

9415

9472

9527

9578

9627

9673

9715

9755

9792

9826

9857

9885

9910

9932

9951

9966

9979

9989

9996

9999

7120

7242

7361

7478

7593

7705

7815

7923

8028

8131

8231

8329

8425

8517

8607

8695

8780

8862

8942

9018

9092

9164

9232

9298

9361

9421

9478

9532

9583

9632

9677

9720

9759

9796

9829

9860

9888

9912

9934

9952

9968

9980

9990

9996

9999

7133

7254

7373

7490

7604

7716

7826

7934

8039

8141

8241

8339

8434

8526

8616

8704

8788

8870

8949

9026

9100

9171

9239

9304

9367

9426

9483

9537

9588

9636

9681

9724

9763

9799

9833

9863

9890

9914

9936

9954

9969

9981

9990

9997

1.000

7145

7266

7385

7501

7615

7727

7837

7944

8049

8151

8251

8348

8443

8536

8625

8712

8796

8878

8957

9033

9107

9178

9245

9311

9373

9432

9489

9542

9593

9641

9686

9728

9767

9803

9836

9866

9893

9917

9938

9956

9971

9982

9991

9997

1.000

7157

7278

7396

7513

7627

7738

7848

7955

8059

8161

8261

8358

8453

8545

8634

8721

8805

8886

8965

9041

9114

9184

9252

9317

9379

9438

9494

9548

9598

9646

9690

9732

9770

9806

9839

9869

9895

9919

9940

9957

9972

9983

9992

9997

1.000

7169

7290

7408

7524

7638

7749

7859

7965

8070

8171

8271

8368

8462

8554

8643

8729

8813

8894

8973

9048

9121

9191

9259

9323

9385

9444

9500

9553

9603

9650

9694

9736

9774

9810

9842

9871

9898

9921

9942

9959

9973

9984

9993

9998

1.000

7181

7302

7420

7536

7649

7760

7869

7976

8080

8181

8281

8377

8471

8563

8652

8738

8821

8902

8980

9056

9128

9198

9265

9330

9391

9449

9505

9558

9608

9655

9699

9740

9778

9813

9845

9874

9900

9923

9943

9960

9974

9985

9993

9998

1.000

0’

0.00

6’

0.10

12’

0.20

18’

0.30

24’

0.40

30’

0.50

36’

0.60

42’

0.70

48’

0.80

54’

0.90

169

Subrahmanyan Chandrasekhar

Success and achievement were not only in the genes of Subrahmanyan

Chandrasekar, the Nobel Prize winner, for Physics in 1983, but also in the

undeterred spirit of determination. Born on 19th October 1910, in Lahore,

he was inspired by his Paternal uncle Sir C.V. Raman. He graduated from

Presidency college and did his higher studies

at the Trinity College, England in 1930.

In 1936, he joined the staff at the University

of Chicago and was honoured as

Morton D Hull distinguished service

professor of the university in 1952. In 1930,

he proved that a star of a mass greater than

1.4 times that of the sun, ends its life

by collapsing into an object of enormous

density. This limit is known as Chandrasekhar’s limit. He had published 400

papers. Theory of radioactive transfer, quantum theory of the negative ion

of hydrogen (1943 - 1950), Mathematical black holes (1971 -1983)

and Theory of Colliding gravitational waves are a few important research

work carried out by him.

He remained active until the evening of his life and published, “Newtons

Principia for the common Reader (1995)”. He died on 21st August 1995 in

Chicago, Illinois, USA. He has left behind his paths, to be treaded upon by

the young challenging minds.

“Seek knowledge like a sinking star beyond the utmost bound ofhuman thought.” Strive ceaselessly success shall crown you gloriously.

Subrahmanyan Chandrasekhar

170

Chandrasekhara Venkataraman

The life and achievements of the illustrious Indian Physicist and Nobel Prize

Awardee, C.V. Raman are an orifice for today’s generation to explore greater

depths, in the field of science, at large and Physics in particular. C.V. Raman was

bron on November 7, 1888 in Tiruchirapalli. He graduated himself in Physics and

English in the highest distinctions. He won the Nobel Prize for his work on the

scattering of light and for the discovery of the

Raman effect. He was the first Indian

scholar to receive the Nobel Prize with an

Indian Education. In 1934, he was the

Director of the Indian Institute of Science,

Bangalore and became the first National

Professor of Independent India. He had

published journals and delivered lectures

about Dynamical Theory of the motion of

Bowed strings (1914) Aspects of Science

(1948) and Physical Optics (1959). In 1949

he established the Raman Research Institute in Banglore. He was

knighted in 1929 and received the Bharat Ratna in 1954. India celebrates

National Science Day on the 28th February to commemorate the discovery of

the Raman effect. Raman died at the age of 82 on November 21, 1970.

A perusal of the lives of these Indian Scientists will certainly remind

you that your progenitors have paved your way to scale greater

scientific heights.

Sir C.V. Raman

1

STD : X Model question paperScience Paper – I

Physics Max. marks : 50 Time : 1 hrs

I. Choose the correct answer and write it against the question numberin your answer book. (10 x 1 = 10)

1. A man carrying a cement bag on his back up a slope will

(a) lean backward (b) lean forward

(c) walk straight (d) lean towards his left.

2. The wavelength of Chennai A broadcasting station is 60 m. At whatfrequency does Chennai A broadcast?(a) 3000 kHz (b) 4000 kHz(c) 5000 kHz (d) 6000 kHz

3. The input voltage of a transformer whose turns ratio 40 is 6 volt. Theoutput voltage is(a) 6 V (b) 40 V(c) 120 V (d) 240 V

4. Which of the following is based on electromagnetic induction ?

(a) transformer (b) galvanometer

(c) loudspeaker (d) motor

5. The unit of stress is(a) N m (b) N m–1

(c) N m–2 (d) N s–1

6. A chain reaction is possible when the mass of the fuel is greater than(a) proton mass (b) neutron mass(c) electron mass (d) critical mass

14

2

7. In a Nuclear reactor the fissionable material is

(a) (b) (c) (d)

8. Natural radioactivity occurs in elements of atomic number greater than(a) 28 (b) 82(c) 52 (d) 40

9. The objects which are found between orbits of Mars and Jupiter are(a) meteorites (b) asteroids(c) comets (d) meteors

10.The ratio of the velocities with which two galaxies move away from theearth is 2:3. The ratio of their distances is(a) 3: 2 (b) 2: 3(c) 4 : 9 (d) 9 : 4

II. Answer any five of the following questions in one or two sentences:( 5 x 2 = 10)

11. Define angular momentum.

12.What is the energy of a photon of frequency 7.5 x 1014 Hz ?

13.The magnetic flux linked with a coil changes from 0.3 Wb to zero in1.2 second. Calculate the induced emf.

14.State Fleming’s left hand rule.

15.Why does a person standing in a railway platform tend to fall towards themoving train?

16.How does the surface tension vary with temperature?

17.Find the nuclear radius of 13

Al 27.

18.What is the principle involved in the production of X-rays?

19.State Newton’s universal law of gravitation.

92U

232

92U

235

90Th

234

94pu

235

3

III. Answer any five of the following questions ( 5 x 3 = 15)

20.Compare the motion of a freely falling body with that of a projectile.

21.Derive the relation between linear velocity and angular velocity.

22.Caluculate the momentum of a particle associated with de Broglie wave ofwavelength 2 A.

23.What is photoelectric effect? On what factors does the photoelectric currentdepend?

24.Find the cost of using a 1500W immersion heater and a 700 W electric ironfor 30 minutes per day for 30 days, at the cost of Rs. 2 per unit.

25.Distinguish between streamline flow and turbulent flow.

26.Define reproduction factor. Give its significance.

27.Write any three properties of X-rays.

28.What are meteors and meteorites?

IV. Answer any three of the following questions ( 3 x 5 = 15)

29.A force of 150 N is required to break a 3m long nylon cord. An object ofmass 1.2 kg is fixed to one end of the cord and whirled around. Determinethe maximum speed with which it can be whirled around withoutbreaking the cord.

30.Explain Raman effect and write its applications.

31.List the parts of a DC generator and describe its working with a neatdiagram.

32.Explain the various energies possessed by a liquid that flows through a pipe.

33.Describe Rutherford’s α - particle scattering experiment and write theinferences.

34.Write the uses of Radio - isotopes.

o

138

PRACTICAL

List of experiments

1. Using simple pendulum determine the acceleration due to gravity in thelaboratory.

2. Determine the melting point of wax by plotting cooling curve.

3. Determine the focal length of the given convex lens.

4. Determine the weight of the given solid using the principle of moments.

5. Determine the relative density (specific gravity) of a liquid using a test-tube

float as a constant immersion hydrometer.

6. Verify the first law of transverse vibration of a stretched string using

sonometer.

7. Determine the refractive index of the material of the given glass prism.

8. Determine the specific resistance of the material of a given wire.

139

1. SIMPLE PENDULUM

AimTo determine the acceleration due to gravity using a simple pendulum.

Apparatus requiredA vertical stand with clamps, a metal bob with a hook, an inelastic thread,

split halves of a rubber cork, a metre scale, a stop-clock and two wooden

blocks.

FormulaAcceleration due to gravity g = 4 π2

where π = 3.14

l = length of the pendulum

T = time period of oscillation

ProcedureA metal bob is tied to one end of a light inextensible thread. Its other end

passes between two halves of a split cork held tightly in a clamp of vertical

stand. The length of the simple pendulum is measured from the lower end of

split cork to the centre of the bob. It is noted as ‘l’. The equilibrium position

of the pendulum is noted as the mean position. The bob is drawn to one side

and it is released so that the pendulum oscillates to and fro in a vertical plane.

When the bob crosses the mean position towards one side, the stop clock is

started and it is counted as zero. Next time when the bob crosses the mean

position towards the same side it is counted as one oscillation. Thus the time

taken for 10 oscillations is noted, from which the time period of oscillation is

calculated as T. The experiment is repeated for various lengths and in each

case the time period is found.

lT2

140

The value of l and the corresponding time period T are tabulated. is

calculated for each trial. The mean value of is calculated. The value of

acceleration due to gravity is calculated using the formula

g = 4 π2 by substituting the mean value of

Tabulation - to determine

Trial Length of Time for 10 oscillations Time forNo the second one

pendulum oscillation

lllll cm T sec sec2 cms–2

1

2

3

4

5

ResultThe value of acceleration due to gravity g = ms-2

trial 1 trial 2 mean

lT2

lT2

lT2

T2lllllT2

Mean =lT2

lT2

lllllT2

simple pendulum

141

2. MELTING POINT OF WAX

AimTo determine the melting point of wax by the cooling curve method.

Apparatus required

A hard glass tube or a boiling tube, a thermometer, vessel containing water(water-bath), a bunsen burner, tripod stand, wire gauze, stop clock andwax pieces.

Procedure

A boiling tube with scraped pieces of wax is taken. A sensitive thermometer isintroduced in it. Now the boiling tube is clamped to a retort stand, so that thetube is immersed into a trough containing boiling water. The solid waxgradually starts melting into a liquid. When the solid wax is completely meltedboiling tube is removed from the trough along with the stand. Now the wax isallowed to cool. The temperature of the molten wax is noted down for every30 second till the temperature reaches 50oC.

A graph is drawn with time along the x-axis and temperature along the y-axis.The temperature corresponding to the horizontal portion (BC) of the graph givesthe melting point of wax.

melting point of wax

142

Tabulation

Time (minutes) Temperature (0C)

0

1

.

.

.

.

.

50oC

ResultThe melting point of wax by the cooling curve method = oC

12

Liquid state

Steady temperature

Solid state

Time (minutes)

A

X

Y

B C

D

Meltingpoint of

wax

Graph

Tem

pera

ture

(0 C)

143

3. FOCAL LENGTH OF CONVEX LENS

AimTo determine the focal length of convex lens by

(i) distant object method(ii) plane mirror method(iii) u-v method and(iv) graphical method

Apparatus requiredThe given convex lens, lens stand, white screen, metre scale, an illuminated

wire gauze and a plane mirror.

FormulaFocal length of the convex lens

where u = distance between the lens and the object.v = distance between the lens and the image.

Procedure(i) Distant object methodThe convex lens is mounted on the stand and is kept facing a distant object

(may be a tree or a building). The white screen is placed behind the convex lensand its position is adjusted to get a clear, diminished and inverted image of theobject. The distance between the convex lens and the screen is measured. Thisgives an approximate value of the focal length of the convex lens.

(ii) Plane mirror methodThe given convex lens is mounted on the lens stand and placed in front of the

illuminated wire gauze at the same level. A plane mirror is held close behind thelens and its reflecting side faces the lens. The plane mirror and convex lens are

uvu + v

f =by u - v method

144

moved close to the wire gauze, till a clear image is obtained on the box by theside of the wire gauze. The distance between the lens and illuminated wire gauzegives the focal length of the convex lens

Focal length of the Convex lens - u-v method

f

Distant object method

Plane mirror method

fM L

145

(iii) U-V method

The convex lens is mounted on the stand and placed in front of the illuminated

wire gauze at a certain distance ‘u’ from the wire gauze. Two values of ‘u’ are

chosen between f and 2f of the lens and the other two values of u are chosen

above 2f. A screen is placed on the other side of lens and its distance from the lens

is adjusted to get a clear image. The value of ‘u’ less than 2f will produce an

enlarged image and that greater than 2f will produce a diminished image. The

distance between the lens and the screen is taken as ‘v’ and it is measured for each

experimental value of ‘u’.

The focal length of the convex lens is calculated using the formula,

(iv) Graphical method

f = uvu + v

u-v graph

A graph is drawn taking the same origin and the same scale with‘v’ along the y-axis and u along the x-axis. The bisector drawn to the angle atorigin cut the u – v graph at C. From C perpendiculars are drawn to both theaxes.

C

A

B

U

V

146

The intercepts OA along x – axis and OB along y – axis are noted.The focal length of the lens is calculated as

ObservationThe focal length of the convex lens by

(i) Distant object method f = _______ x 10-2 m(ii) Plane mirror method f = _______ x 10-2 m

Tabulation

U-V method

Trial No Nature of imageObject Image

distance distanceu cm v cm

1 Real, inverted and

2 magnified u < 2f

3 Real, inverted and

4 diminished u > 2f

Result mean f =

The focal length of the convex lens by

(i) distant object method f = ________ x 10–2 m

(ii) plane mirror method f = ________ x 10–2 m

(iii) u – v method f = ________ x 10–2 m

(iv) graphical method f = ________ x 10–2 m

f = uvu + v

focal length

OB2

OA2

f = =

cm

147

4. PRINCIPLE OF MOMENTS

AimTo determine the weight of a given solid using principle of moments.

Apparatus requiredMetre scale, slotted weights, the given solid of unknown weight and a

knife-edge fixed on a vertical stand.

Principle

When a rigid body is in equilibrium under the action of number of parallelforces, then the sum of the clockwise moments is equal to the sum of theanticlockwise moments.

Formula

w1 =

w2 d

2

d1

w1 = unknown weight.

w2 = weight suspended (slotted weight).

d1 = distance between the centre of gravity and the unknown weight.

d2 = distance between the centre of gravity and the slotted weights.

ProcedureThe metre scale is balanced on the knife-edge of the vertical stand so that it

is horizontal. The balancing point is its centre of gravity. The solid whoseweight is to be found is suspended at a fixed distance, say 25 cm from thecentre of gravity by means of a twine. The slotted weights (w

2) are suspended

on the other side of the knife-edge with the help of a twine. This position isadjusted till the scale attains horizontal position. The distance between thepoint of suspension of the known weights (slotted weights) and the centre ofgravity is measured as d

2.

148

If w1 is the unknown weight of the solid, then by the principle of moments

w1 d

1 = w2

d2

w1 =

w2 d

2

d1

Thus the unknown weight of the solid w1, is calculated. The experiment is

repeated for different values of the known weight w2. The calculations are

done for every trial. The average value of w1 is found. This gives the weight

of the given solid.

Principle of moments

d1 d

2

Unknown weight W1

slotted weights W2

149

Tabulation

Trial Known weightWeight of the solid

No w2

w1 =

w2 d

2

d1

1

2

3

4

5

mean w1 =

ResultThe weight of the given solid using principle of

moments w1= x 10-3 kg wt

theunknown weight

d1 cm

theslotted weights

d2

cmg wt g wt

Distance of Distance of

150

5. CONSTANT IMMERSION HYDROMETER

AimTo determine the specific gravity or relative densityof a liquid using a test

tube float as a constant immersion hydrometer.

Apparatus requiredA test tube float, lead shots or sand, two tall jars containing water and the given

liquid and a spring balance.

Formula

Specific gravity or relative density of the liquid=weight of the float in liquidweight of the float in water

=w

2

w1

ProcedureThe test tube is made to float in water vertically to a certain depth by adding

sand or lead shots. The depth of immersion of the float in water is noted as ‘h’.The float should not touch the bottom or sides of the jar and there should be no airbubbles sticking to its sides. The test tube is taken out and its outer surface iswiped off. Then it is suspended from the hook of a spring balance and its weightin water is noted as w

1 g wt. Then it is allowed to float to the same depth ‘h’ in

the given liquid by adding or removing lead shots. Its outside is wiped dry andagain its weight in the liquid is found using a spring balance as w

2 g wt.

The experiment is repeated by varying the depth of immersion ‘h’. Thereadings are tabulated. The specific gravity of the liquid using the test tubefloat as a constant immersion hydrometer is calculated using the formula,

=w

2

w1

Specific gravity or relative density of the liquid

151

Tabulation

Trial

Depth of the Weight of the float Specific gravity of

No

immersion of the liquid

1

2

3

4

5

ResultThe specific gravity of the liquid using the test tube float as a constant immersion

hydrometer = _______ (no unit)

in waterw

1 g wt

in liquidw

2 g wt

floath cm

mean w

2 = w1

w2

w1

test tube float

152

6. SONOMETER

AimTo verify the first law of transverse vibrations in stretched string using

sonometer.

Apparatus requiredSonometer, tuning fork, rubber hammer, slotted weights and paper rider.

LawThe frequency of the vibration is inversely proportional to length of the

vibrating segment.

Formulan l = constant (tension of the wire is kept constant).

where n = frequency of the resonating wire.(i.e. the frequency of the tuning fork)

l = length of the resonating wire.

Procedure

The sonometer wire is kept under tension by a suitable load, say 3 kg.A small paper rider is placed on the wire between the movable bridges.A tuning fork of known frequency ‘n’ is excited by striking it gently witha rubber hammer. It is then placed with the stem on the sonometer box. Nowthe wire segment between the movable bridges will vibrate. By moving onemovable bridge gently, the length of the vibrating segment is adjusted till thepaper rider flutters violently and is thrown off. (If the tuning fork stops vibrating,it is again excited and placed on the sonometer box). At this stage resonanceis said to occur (i.e.) the frequency of the tuning fork is equal to thefrequency of the vibrating segment.

153

Tabulation

S. No. Frequency of the tuning fork Resonating lengthn l

n l

1

2

3

Result

n lllll = constantThus it is proved that the frequency is inversely proportional to the vibrating

length.

The length of the wire between the two movable bridges gives the resonatinglength ‘l’. Keeping the tension constant, the experiment is repeated with tuningforks of different frequencies and the corresponding resonating lengths of thewire are found out as before. The values are tabulated and ‘nl ’ is found tobe a constant

Hz

l

cm

sonometer

x 10–2 Hz m

154

7. REFRACTION THROUGH GLASS PRISM

AimTo determine the refractive index of the material of the given glass prism.

Apparatus requiredGlass prism, drawing board, sheets of white papers, instrument box, pins and

board pins.

Formula

µ =

whereµ = refractive index of the glass prismA = angle of the prism.D = angle of minimum deviation.

Procedure

A sheet of white paper is fixed on a drawing board. The glass prism is placedon the paper and its outline ABC is drawn. The prism is removed. The point D ismarked on the side AB (near its midpoint). A normal ED is drawn at D using aprotractor. A line RD is drawn at an angle say 30o to ED. This ∠ RDE=30o is theangle of incidence and RD is the incident ray. Two pins P and Q are placedvertically on the line RD. Now the prism is placed on its outline. On lookingthrough the face AC of the prism, the images of P and Q are seen. Two pins F andG are fixed vertically at two different positions, so that they lie in line with theimages of P and Q. Now MN is the emergent ray. Produce the incident ray andthe emergent ray backward into the outline of the prism. The two rays meet at apoint O. The external angle formed at O is the angle of deviation ‘d’. The experimentis repeated for different angle of incidence.

Sin A2

Sin A + D

2(no unit)

155

i-d graphA graph is plotted taking ‘i’ along x - axis and ‘d’ along ‘y’ - axis. It is shown

below:

From the graph, it is seen that as the angle of incidence increases the angle ofdeviation decreases, reaches a minimum and then increases. This minimum valueis called angle of minimum deviation, D.

Substituting the values of angle of the prism (A = 60o) and the angle ofminimum deviation (D), the refractive index of the material of the prism canbe calculated using the formula,

Sin A2

Sin A + D

2µ =

d

i

Y

i-d curve

156

Tabulation

S. No. Angle of incidence Angle of deviationi d degree

1 30o

2 35o

3 40o

4 45o

5 50o

ResultThe refractive index of the material of the prism µ = _________ (no unit)

Refraction through a glass prism

R

E

INCIDENT RAY

O

F

G

N

D M

157

8. SPECIFIC RESISTANCE OF A WIRE

AimTo determine the specific resistance of the material of the wire (given the radius

of the wire and the length of the wire).

Apparatus requiredA battery eliminator, ammeter, voltmeter, key, rheostat, experimental wire and

connecting wires.

Formula

Resistance of the wire R= V I

Specific resistance of the wire S = πr2Rl

S = specific resistance of the wireR = resistance of the given wirer = radius of the given wirel = length of the given wire

ProcedureConnect the battery eliminitor, ammeter the given wire, rheostat and key in

series. The voltmeter is connected in parallel connection across the given wire.

The circuit is closed and the rheostat is adjusted such that a constant current

flows through the given coil of wire. The current is noted as ‘I’ from the ammeter

and the potential difference across the wire V is noted from the voltmeter.

The value v gives the resistance of the wire. The experiment is repeated for

different values of the current.

The average value of V gives the resistance of the wire R.

I

I

158

The radius of the wire is given as ‘r’ metre. The length of the wire is given as ‘l’ metre.

Specific resistance of the material of the wire is calculated using the formula

S = πr2 R

l

TabulationTo determine the resistance of the wire

Trial No Ammeter reading Voltmeter reading ResistanceI ampere V volt R = V I ohm

1

2

3

4

5

mean R =

159

Observations

length of the wire l = x 10-2 m

radius of the wire r = x 10-2 m

resistance of the wire R = ohm

Specific resistance S = π r2 R ohm m

l

ResultSpecific resistance of the material of the wire S = _______ ohm m.

160

Important Constants

Name Symbol Value

Speed of light c 3 × 108 m s–1

Gravitational constant G 6.67 × 10–11 Nm2 kg–2

Acceleration due to gravity g 9.8 ms–2

Permittivity of free space εο 8.854 × 10–12 C2 N–1 m–2

Permeability of free space µο 4π × 10–7 henry / metre

Charge of an electron e 1.602 × 10–19 C

Mass of an electron me 9.1 × 10–31 kg

Mass of a proton mp 1.67 × 10–27 kg

Mass of a neutron mn 1.675 × 10–27 kg

Planck’s constant h 6.625 × 10–34 Js

161

STD : X SCIENCE PAPER – I Marks : 25 Time : 1 hrs

SECTION A : PHYSICS PRACTICAL

MODEL QUESTION PAPER

1. Using simple pendulum determine the acceleration due to gravity in thelaboratory. ( atleast 4 readings)

(or)Verify the first law of transverse vibration of a stretched string usingsonometer. ( atleast 3 readings)

2. Determine the melting point of wax by plotting cooling curve.(or)

Determine the refractive index of the material of the given glass prism.( atleast 5 readings)

3. Determine the focal length of the given convex lens by u-v method. Verifyby plane mirror method. ( atleast 5 readings)

(or)Determine the relative density (specific gravity) of a liquid using a test-tubefloat as a constant immersion hydrometer. ( atleast 4 readings)

4. Determine the weight of the given solid using the principle of moments.( atleast 4 readings)

(or)Determine the specific resistance of the material of a given wire.( atleast 4 readings)

5. Determine the relative density (specific gravity) of a liquid using a test-tubefloat as a constant immersion hydrometer. ( atleast 4 readings)

(or)Verify the first law of transverse vibration of a stretched string usingsonometer. ( atleast 3 readings)

6. Determine the specific resistance of the material of a given wire.( atleast 4 readings)

(or)Determine the refractive index of the material of the given glass prism.( atleast 5 readings)

12

162

7. Verify the first law of transverse vibration of a stretched string usingsonometer. ( atleast 3 readings)

(or)Determine the melting point of wax by plotting cooling curve.

8. Determine the weight of the given solid using the principle of moments.( atleast 4 readings)

(or)Determine the melting point of wax by plotting cooling curve.

9. Determine the focal length of the given convex lens by u-v method. Verify byu-v graph method. ( atleast 5 readings)

(or)Determine the specific resistance of the material of a given wire.( atleast 4 readings)

10. Determine the focal length of the given convex lens by u-v method. Verify by u-v graph method. ( atleast 5 readings)

(or) Using simple pendulum determine the acceleration due to gravity in the laboratory. ( atleast 4 readings)

Scheme for Evaluation

Record – 5 marks

Formula – 3 marks

Procedure – 3 marks

Tabulation – 2 marks

Observation – 5 marks

Calculation – 5 marks

Result – 2 marks

Total – 25 marks

MCQ VSA SA LA MCQ VSA SA LA MCQ VSA SA LA MCQ VSA SA LATOTALOBJECTIVES

KNOWLEDGE UNDERSTANDING APPLICATION SKILL

1(1) 1(2) 1(3) 1(3) 1(5) 14

1(3) 1(5) 1(1) 1(2) 1(3) 14

1(2) 1(1) 1(5) 1(1) 1(2) 1(3) 14

1(1) 1(3) 1(2) 1(5) 1(2) 13

1(5) 2(1) 1(3) 1(2) 12

1(1) 1(2) 1(3) 1(5) 11

1(1) 1(2) 1(3) 1(1) 7

4 8 12 5 3 2 3 10 2 6 10 3 6 6 5 85

BLUE PRINT

MCQ - Multiple Choice Question; VSA - Very Short Answer; SA - Short Answer; LA - Long Answer.

Note: Figures within brackets indicate marks & figures outside the brackets indicate the number of question.Scheme of sections : MCQ VSA SA LAScheme of options : 10/10 5/9 5/9 3/6

Science Theory Paper - Physics Maximum Marks - 50 Time : 1¼ Hours

Mechanics

Light

Electricity

Properties ofMatter

Modern physics

X - rays &radioactivity

Universe

4